Simple Harmonic Motion – Questions and solutions

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Chemical Kinetics (Subjective Unsolved)

The nuclei of two isotopes of same substance A-236 & A-234 are present in 4:1 ratio in an ore obtained from some other planet. The half-lives are 30 minute and 60 minute respectively. Both isotopes are a-emitter. Calculate after how much time their activity will be identical and also calculate the time required to bring the ratio of their atoms 1:1.

180 minute and 1200 minute

To investigate the beta-decay of Mg-23 radionuclide, a counter was activated at the moment t = 0, it registers N₁ beta particles by a moment t₁= 2 sec and by a moment
t₂ =3t₁, the number of registered particles was 2.44 times greater. Find the mean time of the given nuclei.

t_av= 15.9 hours

For the following sequential reaction

a) Find out the time for maximum rate of formation of B. Given that K₁ = 1.5 ´ 10^–4sec^–1 and K₂ = 1.5 ´ 10^–7 sec^–1.
b) Find out the concentration of C, when the initial concentration of A is 1.5 M and given that K₁ = 1.2 ´ 10^–5 sec^–1 and K₂ = 1.1 ´ 10^–2 sec^–1, t = 1 day

a) 25.59 days, (b) 1.325 M

A solution contains a mixture of isotopes I₁ (t_1/2 = 14 days) and I₂ (t_1/2= 25 days). Total activity is 1 curie at t = 0. The activity reduces by 50% in 20 days, then find
a) the initial activity I₁ and I₂
b) The ratio of their initial no. of nuclei.

a) = 0.3669, = 0.6331
b) = 0.3245

In presence of an acid, N-chloroacetamide slowly changes into p-chloroacetamide. The progress of the reaction can be measured by titrating iodine liberated with Na₂S₂O₃ solution. The results were obtained as follows:

Time (in hours) 1 2 6 ¥

Amount in p-chloroacetamide (in ml) 13 22.5 39.3 45

Find the fraction of N-chloroacetamide decomposed after 3 hours.

0.643

What amount of heat is evolved by a curie of Rn (an a-emitter) in
a) two hour
b) its mean-life?

Given that kinetic energy of an a-particle is 5.5 MeV and l – 2 ´ 10^–6 sec^–1 for reaction.

a) 11.45 J, (b) 1.03 ´ 10⁴J

Atomic Structure (Subjective Unsolved)

If the radius of the first Bohr orbit in a hydrogen atom can be determined with an uncertainty of 1% of its actual value, what will be the uncertainty in the velocity of electron? Compare the value with the velocity of electron in the first orbit. Given that
m_e = 9.1 ´ 10^–31 kg and E₀ = 8.854 ´ 10^–12 C²N^–1 m^–2.

2.19 ´ 10⁶ms^–1

The wavelength of K_a characteristic X-ray of iron and potassium are 1.931 ´ 10^–8 and 3.737 ´ 10^–8 cm respectively. What is the atomic number and name of the element of which characteristic K_a wavelength is 2.289 ´ 10^–8 cm.
Atomic number = 24, Cr
An hydrogen like atom (atomic number = Z) is in a higher excited state of quantum number n. This excited atom make a transition to the first excited state by successively emitting by two photons of energies 10.20 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy 4.25 eV and 5.95 eV respectively. Determine the value of n and Z.
n = 6, Z = 3
Light of wavelength 2000Å falls on an aluminium surface (work function of Al is 4.2eV). Calculate (a) the kinetic energy of the slowest and fastest emitted photon electrons. (b) stopping potential. (c) Cutt-off wavelength for Al.
a) 2.0 eV; (b) 2V; (c) 2970Å
According to Mosely, frequency of X-ray and the charge present on the nucleus of atom of the element has the following relation

= a( Z – b)

Where n = frequency

Z = atomic number

a, b are constant

a) If the straight line is at an angle of 45° with intercept 1 on axis, calculate the frequency n when atomic number Z = 50.
b) What is the atomic number when n = 2500 sec^–1.

a) n = 2401 sec^–1
b) Z = 51

An electron in a H-atom in its ground state absorbs 1.50 times as much as energy as the minimum required for its escape (13.6eV) from the atom. Calculate the wavelength of the emitted electrons.

4.70 Å

Atomic Structure (Objective)

Choose the correct alternative only one is correct.

The nucleus of an atom is located at x = y = z = 0. If the probability of finding an s-orbital electron in a tiny volume assumed x = a, y = z = 0 is 1 ´ 10^–5, what is the probability of finding of the electron in the same sized volume around x = z = 0, y = a?

(A) 1 ´ 10^–5 (B) 1 ´ 10^–5a

What is the probability at the second site if the electron were in a p_z orbital for the data given in question number (1)

(A) 1 ´ 10^–5 (B) 2 ´ 10^–5

Following ions will be coloured if Aufbaun rule is not followed

(A) Cu⁺² (B) Fe⁺²

If wavelength is equal to distance travelled by the electron in one second, then

(A) (B)

Number of photons of light of wavelength 4000Å required to provide 1.00J of energy is

(A) 2.01 ´ 10¹⁸ (B) 12.01 ´ 10³¹

When a certain metal was irradiated with a light of frequency 3.2 ´ 10¹⁶ Hz the photoelectrons emitted had twice the kinetic energy as did photo electrons emitted when the same metal was irradiated with light frequency 2.0 ´ 10¹⁶Hz. Hence threshold frequency is

(A) 1.6 ´ 10¹⁶ Hz (B) 0.8 ´ 10¹⁶ Hz

The radial distribution curve of 2s sub-level consists of ‘a’ nodes, a is

(A) 1 (B) 3

Uncertainty in position and momentum are equal. Uncertainty in velocity is

(A) (B)

If the radius of first Bohr orbit of H-atom is x, then de-Broglie wavelength of electron in 3^rd orbit is nearly:

(A) 2px (B) 6px

If the shortest wavelength of H-atom in Lyman series is x, then the longest wavelength in Balmer Series of He⁺is

(A) (B)

Solved Objective

Solution 11: Since s-orbital is spherically symmetrical and has equi-distance from the nucleus. Hence the probability is identical

\ (A)

Solution 2: P₂ has a node at z = 0, hence probability of finding of e^– in this volume is zero

\ (D)

Solution 3: Ion with unpaired electrons in d or f-orbitals will be coloured

\ (D)

Solution 4: Since, the distance travelled in one second by velocity = vc_m = l

Q l =

\ l = Þ l =

\ (D)

Solution 5: Q E = nhn25 where n = no. of photons

\ E =

\ n = = 2.01 ´ 10¹⁸

\ (A)

Solution 6: Q hn = hn₀ + KE

\ KE = h(n – n₀)

Again 2(KE)₁ = (KE)₂

2h(n₁– n₀) = (n₂ – n₀) ´ h

Þ 2(n – n₀) = (n₂ – n₀)

\ Þ 2n₁ – n₂= n₀

Since n₁ = 2 ´ 10¹⁶ Hz

n₂ = 3.2 ´ 10¹⁶ Hz

\ n₀ = 8 ´ 10¹⁵ Hz

\ (C)

Solution 7: Since the radial nodes

= n – l – 1

Hence a = 2 – 0 – 1 = 1

\ (A)

Solution 8: Given Dp = Dx

Since Dp ´ Dx »

(Dp)² »

Dp »

\ Dv =

\ (C)

Solution 9: Q mvr₃ =

mv =

Þ mv =

\ l = = 6px

\ (B)

Solution 10: For the shortest wave length of Lymann series in H-atom

Þ = R_H

Let the longest wavelength of Balmer series is l

\ l_max =

\ (A)

Atomic Structure (Unsolved Problems Objective)

Wave function vs distance from nucleus graph of an orbital is given below:

The number of nodal sphere of this orbital is

(A) 1 (B) 2

For an electron in a hydrogen atom the wave function, y is proportional to , where a₀ is the Bohr’s radius. What is the ratio of probability of finding the electron at the nucleus to the probability of finding it at a₀.

(A) e (B) e²

What transition in He⁺ion shall have the same wave number as the first line in Balmer series of H-atom

(A) 3 ® 2 (B) 6 ® 4

The potential energy of the electron in an orbit of H-atom would be

(A) – mv² (B)

An electron is moving with a kinetic energy of 4.55 ´ 10^–25 Joules. What will be the de-Broglie wave length for this electron?

(A) 5.28 ´ 10^–7 m (B) 7.28 ´ 10^–7m

If each orbital can hold a maximum of 3-electrons. The number of elements in 4^th period of periodic table is:

(A) 48 (B) 57

The no. of orbital in a sub-shell is equal to

(A) n² (B) 2l

Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of principal quantum no. n.

(A) a (B) b

The ratio of the energy of the electron in ground state of hydrogen to that of the electron in the first excited state of Be⁺³is

(A) 1 : 4 (B) 1 : 8

The electronic transition from n = 2 to n = 1 will produce the shortest wavelength in

(A) H-atom (B) D-atom

The wave number of first line of Balmer series of hydrogen is 152,00 cm^–1. The wavelength of first Balmer line of Li⁺²ion is
(A) 15,200 cm^–1 (B) 60,800 cm^–1
(C) 76,000 cm^–1 (D) 1,36,8000 cm^–1
The dissociation energy of H₂ is 430.53 kJ mol^–1. If H₂ is dissociated by illumination with radiation of wavelength 253.7 nm. The fraction of the radiant energy which will converted into kinetic energy is given by
(A) 86% (B) 2.33%
(C) 3% (D) 100%
The orbital angular momentum of an electron in 2s-orbital is
(A) (B) zero
(C) (D)
Photoelectric emission is observed from a surface for frequencies v, and v₂ of the incident radiation (where v₁ > v₂). If the maximum kinetic energy of the photoelectrons in two cases are in the ratio of 1:k, then the threshold frequency v₀ is given by
(A) (B)
(C) (D)
The difference between nth and (n + 1)th Bohr’s radius of H-atom is equal to its (n – 1)th Bohr’s radius. The value of n is
(A) 1 (B) 2
(C) 3 (D) 4
The number of nodal planes in a P_x= orbital is
(A) One (B) two
(C) Three (D) Zero
Chemical Kinetics (Solved Objective)
Problem 1 For the reaction 2NO + Br₂ ¾® 2NOBr, the following mechanism has been given
fast
NO + Br₂ NOBr₂
NOBr₂ + NO 2NOBr
Hence rate law is;
(A) k[NO]²[Br₂] (B) k[NO][Br₂]
(C) k[NOBr₂] [NO] (D) k[NO][Br₂]²
Solution: For the reaction
NOBr₂ + NO 2NOBr
\ rate = k₁[NOBr₂] [NO] …(1)
But from the reaction
NO + Br₂ NOBr₂
K_equilibrium =
Þ [NOBr₂] = K_equilibrium [NO] [Br₂] …(2)
Putting equation (2) in equation (1)
rate = k₁ k_equilibrium [NO]²[Br₂]
= k[NO]²[Br₂], where k = k₁ k_equilibrium
\(A)
Problem 2: Following is the graph between (a – x)^–1 and time for second order reaction, Q = tan^–1 (0.5), OA = 2L mol^–1.
Hence rate at the start of the reaction is:
(A) 25 L mol^–1 min^–1 (B) 0.5 mol lit^–1 min^–1
(C) 125 L mol^–1 min^–1 (D) 1.25 L mol^–1 min^–1
Solution: For second order reaction,
Kt =
\ (a – x)^–1 = Kt +
\ k = tanQ (\ tanQ = 0.5) from question
\ K = 0.5 mol lit^–1 min^–1
and = 2 L mol^–1
\ (C)
Problem 3: Half-life period in Q.N. = (2), above is
(A) 386 min (B) 4 min
(C) 16 min (D) 2 min
Solution: t_1/2 (for 2^nd order reaction) =
\ (B)
Problem 4: The rate constant for the reaction
2N₂O₅ ¾® 4NO_2(g) + O_2(g) is 3 ´ 10^–5 sec^–1. If the rate is 2.4 ´ 10^–5 mol lit^–1 sec^–1, the concentration of N₂O₅ in mol L^–1 is;
(A) 4 (B) 1.2
(C) 04 (D) 0.8
Solution: The decomposition of N₂O₅ is a first order reaction
\ rate = k[N₂O₅]¹
4 ´ 10^–5 = 3 ´ 10^–5[N₂O₅]
\ [N₂O₅] = = 0.8 mol L^–1
\ (D)
Problem 5: Following is the graph between logT₅₀ and log a (where a = initial concentration) for a given reaction at 27°C. Hence order is;
(A) 0 (B) 1
(C) 2 (D) 3
Solution: T₅₀ µ a^(1–n) where n = order of reaction
Þ T₅₀ = ka^{(1 – n)}
or, logT₅₀= logk + (n – 1)log a
from question Q = tan^–1(m)
Þ n – 1 = 1
Þ n = 2
\ (C)
Problem 6: For the reaction
is equal to
(A) k₁(a – y) – k₂(a – y) (B) k₂(a – y) – k₁ (a– y)
(C) k₁(a – y) + k₂(a – y) (D) –k₁(a – y) – k₂(a – y)
Solution: = k₁(a – y) + k₂(a – y)
\ = – k₁(a – y) – k₂(a – y)
\ (D)
Problem 7: There are two radionuclei A & B A is an a-emitter and B is a b-emitter, their disintegration constant are in the ratio of 1:2. What should be the number of atoms of two at time t = 0. So that probability of getting of a and b-particles are same at time t = 0
(A) 2 : 1 (B) 1:2
(C) 1:4 (D) 4:1
Solution: Since the rate of disintegrations are same
l_AN_A = l_B×N_B
Þ =
\ (A)
Problem 8: At radioactive equilibrium, the ratio between the atoms of two radioactive elements A and B was found to be 3.1 ´ 10⁹ : 1 respectively. If T₅₀ of the element A is 2 ´ 10¹⁰ years, then T₅₀ of the element B is
(A) 2 ´ 10⁹ years (B) 6.45 years
(C) 3 ´ 10⁸ years (D) None
Solution: At radioactive equilibrium
t_50(B) =
= = years
\ (B)
Problem 9: The no. of b-particles emitted during the charge
_ax^C ¾® _dy^b is
(A) (B) + c
(C) – a (D) – c
Solution: _ax^c ¾® _dy^b + m × ₂He⁴ + n × _–1b⁰
Q
Hence a = d + 2m – n
\n = d + 2m – a
= – a
\ (C)
Problem 10: One mole of A present in a closed vessel undergoes decay as
_zA^m ¾® _z–4B^{m – 8} + 2 × ₂He⁴
The volume of He collected at NTP after 20 days (t_1/2(A) = 10 days) is
(A) 2 L (B) 22.4 L
(C) 6 L (D) 67.2L
Solution: Q = 2ⁿ where n = no. of half lives
= = 2² Q n = = 2
\ N = mole
\ decayed moles = moles
But moles of He formed = moles
= ´4 L at NTP
= 33.6 L at NTP
\(C)
Chemical Kinetics (Unsolved)
Choose the correct alternate(s) one or, more than one may be correct
A radioactive isotopes x with half-life of 1.37 ´ 10⁹ years decays to y which is stable. A sample of rock from the moon was found to contain both the elements x and y in the ratio 1:7. What is the age of the rock.
(A) 96 ´ 10⁸ years (B) 3.85 ´ 10⁹ years
(C) 11 ´ 10⁹ years (D) 3.06 ´ 10⁹ years
A radioactive mixture containing a short lived species A and short lived species B. Both emitting a-particles at a given instant, emits at rate 10,000 a-particles per minute. 10 minutes later, it emits at the rate of 7000 particles per minute. If half lives of the species are 10 min and 100 hours respectively, then the ratio of activities of A : B in the initial mixture was
(A) 3:7 (B) 4:6
(C) 6:4 (D) None
₉₂U²³⁸ (IIIB) undergoes following emissions:
₉₂U²³⁸
Which is/are correct statements
(A) A will be of IB group (B) A will be of III B group
(C) B will be of IIA group (D) C will be of IIIA group
The electron in a hydrogen atom makes transition from M shell to L, the ratio of magnitude of initial to final acceleration of electron is:
(A) 9:4 (B) 81:16
(C) 4:9 (D) 16:81
On order to determine the volume of blood in an animal without killing it, a 1.00 ml sample of an aqueous solution containing tritium is injected into the animal blood stream. The sample injected has an activity of 1.8 ´ 10⁶ cps (counts per second). After sufficient time for the sample to be completely mixed with the animal blood due to normal blood circulation, 2.00 ml of blood are withdrawn from animal and the activity of the blood sample withdrawn is found to be 1.2 ´ 10⁴ cps. Calculate the volume of the animal blood.
(A) 300ml (B) 200ml
(C) 250 ml (D) 400 ml
A radioactive isotopes is being produced at a constant rate x. Half-life of the radioactive substance is y. After sometimes no. of radioactive nuclei becomes constant, the value of this constant is
(A) (B) xy
(C) (ln2)xy (D)
Number of nuclei of a radioactive substance at time t = 0 are 1000 and 900 at time = 2 sec. Number of nuclei at t = 4s will be
(A) 800 (B) 810
(C) 790 (D) 700
There are two radioactive substances A and B. Decay constant of B is twice that of A. Initially both have equal no. of nuclei. After n-half lives of A, the rate of disintegration of both becomes equal, the value of n is
(A) 1 (B) 2
(C) 4 (D) None
A radioactive nuclide is produced at a constant rate of a-per second. It decay constant is l. If N₀ be the no. of nuclei at time t = 0, then maximum no. of nuclei possible are
(A) (B) N₀ +
(C) N₀ (D) + N₀
For the reaction
R – X + OH^–‑ ¾® R – OH + X^–, the rate expression is given as rate
= 4.7 ´ 10^–5 [R – X] [OH^–] + 0.24 ´ 10^–5 [R – X]
What % of R – X react by S_N2 mechanism when [OH^–] = 0.001 M
(A) 9 (B) 4.7
(C) 8 (D) 4.9
The mechanism of the reaction
2NO + O₂ ¾® 2NO₂ is
k₁
NO + NO N₂O₂ (fast)
k_–1
N₂O₂ + O₂ 2NO₂ (slow)
the rate constant for the reaction is
(A) k₂ (B) k₂×k₁×k_–1
(C) k₂×k₁ (D)
The inversion of cane sugar proceeds with half-life of 600 minutes at pH = 5 for any concentration of sugar. However if pH = 6, the half-life changes to 60 minute. The rate law expression for sugar inversion can be written as
(A) r = k×[sugar]² [H⁺]⁰ (B) r = k× [sugar]¹ [H⁺]⁰
(C) r = k× [sugar]¹ [H⁺]¹ (D) r = k× [sugar]⁰ [H⁺]¹
Rate of chemical reaction is nA ¾® Product, is doubled when the concentration of A is increased four times. If the half-life time of the reaction at any temperature is 16 minutes, then the time required for 75% of the reaction to complete is
(A) 0 minutes (B) 27.3 minutes
(C) 48 minutes (D) 4 minutes
The rate of a chemical reaction generally increases rapidly even for small temperature increase because of rapid increase in the
(A) Collision frequency
(B) Fraction of molecules with energies in excess of the activation energy
(C) Activation energy
(D) Average kinetic energy of the molecule
Rate constant, k = 1.8 ´ 10⁴moles^–1 Ls^–1 and E_a = 2 ´ 10² kJ mole^–1, when T ® ¥, then the value of A is
(A) 8 ´ 10⁴ kJ mol^–1 (B) 1.8 ´ 10⁴ mol^–1L sec^–1
(C) 8 ´ 10⁴mol L^–1 sec^–1 (D) 2.4 ´ 10³ kJ mol^–1 sec^–1
Atomic Structure (Solved Subjective)
Solution 1: \ E = nhn where n = no. of photons
=
= =
= 2.51`3 ´ 10²³ of photons or no. of I₂ molecules
= = 0.417 moles of I₂
Solution 2: Let the no. of photons needed to see the object
Q E = nhn =
\ n =
=
= = 28 (approx)
Solution 3: Q KE = mv²
\
From De-broglie’s Principles
l = …(1)
Putting the value of
h = 6.625 ´ 10^–34 jS
me = 9.11 ´ 10^–31 kg
E. = e.v.
Where charge e^– = 1.9 ´ 10^–19c
Putting all these values in equation (1) and or solving
l = Å
= Å = Å
= 5.69 ´ 10^–10m
\ where number = ´ 10¹⁰ meter^–1
= 1.76 ´ 10⁹m^–1
Solution 4: a) E. of the emitted electrons
= mv², where m = mass of electron
= 9.11 ´ 10^–3 kg
v = velocity of emitted electrons
= 6.4 ´ 10⁴ m/s
\E. = ´ 9.11 ´ 10^–31kg ´ (6.4 ´ 10⁴ m/s)²
= ´11 ´ 10^–31 ´ 6.4 ´ 6.4 ´ 10⁸J
= 186.57 ´ 10^–23J
= 1.86 ´ 10^–21J
b) \ hn = w₀ + (K.E.)_maximum …(1)
Where w₀ = work function = minimum amount of energy required to remove electron.
From equation (1)
= w₀ + (KE)_maximum
Þ = w₀ + 1.86 ´ 10^–21J
Þ8 ´ 10^–21J = w₀ + 1.86 ´ 10^–21J
Þ w₀ = 421.01 Joules
Solution 12: a) Q8gm hydrogen atoms = 1.8 ´ 6.023 ´ 10²³ atoms of H
of atoms present in 3^rd energy level
= 1.8 ´023 ´ 10²³ ´
= 2.927 ´ 10²³ atoms
Similarly no. of atoms present in 2^nd energy level
= 1.8 ´023 ´ 10²³ ´ 0.15 atoms of H
= 1.626 ´ 10²³ atoms of H
b) When all H-atom jumps from energy level 3^rd to ground state, the amount of energy released,
E₁ = 13.6 ´ ´927 ´ 10²³ eV
= 35.834 ´ 10²³ eV
Similarly when all H-atom present in orbit no. of two jumps to ground state, let the amount of energy released
E₂ = 13.6 ´626 ´ 10²³ eV
= 16.585 ´ 10²³ eV
Since energy is additive, so, total energy released
E_T = E₁ + E₂
= 51.969 ´ 10²³ eV
= 83.15 ´ 10⁼⁴ J
Chemical Kinetics (solved)
Solution 1: The minimum b-activity required is 346 b-particles per minute. Since the experiment has to be completed in 6.909 hours. Moreover, we are asked to determine the minimum amount of required to carryout the experiment
Rate of emission of b-particles = 346 particles per minutes
Total b-particle required in 6.909 hours
= 346 ´909 ´ 60 = x
= 143244 = x
Q lt = 2.303 log
= 2.303
Solving, a = 2.07 ´ 10⁶ b-particles
43 ´ 10^–18 moles =
402 ´ 10^–16gm = 3.48 ´ 10^–18 ´ 99 gm
Solution 6: a) Since = 2ⁿ
Where n = no. of half lives
= = 4
\ = 2⁴ = 16
\ Remaining no. of atoms of tritium, after 49 years
N = ´ N₀ = ´ 8 ´ 10^–18 = 5 ´ 10^–19
But atom remains unchanged = 1
Hence ratio of to normal = 1
= 5 ´ 10^–19 : 1
b) 10 gm H₂O = moles of H₂O
= moles of H-atom
= ´023 ´ 10²³ H-atom
= ´023 ´ 10²³ ´ 8 ´ 10^–18Tritium atoms
N₀ = 5.35 ´ 10⁶ tritium atoms
Again
N = N₀ ´ 2^–n= 5.35 ´ 10⁶ ´ tritium atoms
N = 0.3344 ´ 10⁶ tritium atoms
Solution 10: From question
₈₄Po²¹⁸ ₈₂Pb²¹⁴ ₈₃X²¹⁴
Let ₈₄Po²¹⁸ changes into ₈₂Pb²¹⁴ with rate constant k₁
and similarly ₈₂Pb²¹⁴ changes into ₈₃X²¹⁴ with rate constant k₂
Therefore k₁ = = = 0.227 min^—1
k₂ = = = 0.2585 min^–1
For maximum conc. of ₈₂Pb²¹⁴,
(k₂ – k₁)t = ln = ln =0.1299
Solving,
t = = 4.12 min
Solution 12: Since, from unit of rate constant, i.e. sec^–1, it is clear that it is the 1^st order reaction
k(overall) = 1.52 ´ 10^–4s^–1
Let B is forming with rate constant = k₁
and C is forming with rate constant = k₂
For parallel path, k = k₁ + k₂
From question,
= 11.2% Þ k₁ = 11.2% ´ k
= 0.112 ´52 ´ 10^–4s^–1
= 0.17024 ´ 10^–4s^–1
= 1.7024 ´ 10^–5s^–1
Therefore = 88.8%
\ k₂ = 88.8% ´ k
= 0.888 ´52 ´ 10^–4 s^–1
= 1.349 ´10^–4s^–1
Solution 15: Since = q – A, where A = lN
\ = q – lN
\
Integrating both sides with proper upper limit and lower limit
= = [t – 0]
on solving, = lt
´ t
On solving, t = 8.24 ´ 10⁵sec
= 9.54 days

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