Dual Nature of Matter and Radiation Test

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Dual Nature of Matter and Radiation Test - 9

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Weekly Quiz Competition

Question 1
1 / -0.25

In a hydrogen atom, the electron is in nth excited state. It may come down to second excited state by emitting ten different wavelengths. What is the value of n

A
6

B
7

C
8

D
5

Solution
Question 2
1 / -0.25

Difference between nth and (n + 1)th Bohr 's radius of `H 'atom is equal to it 's (n ^_ 1)th Bohr 's radius. The value of n is

A
1

B
2

C
3

D
4

Solution

We know that r_n ∝n²
Given: r_n+1 −r_n =r_{n −1}
∴(n+1)² −n² =(n −1)²
n=4
Question 3
1 / -0.25

The electron in a hydrogen atom makes transition from M shell to L. The ratio of magnitudes of initial to final centripetal acceleration of the electron is

A
9 : 4

B
81 : 16

C
4 : 9

D
16 : 81

Solution

According to Bohr 's atomic theory,
radius of nth orbit, r ∝n² /Z
velocity of electron in nth orbit, v ∝Z/n
then acceleration of electron in nth orbit, a = v² /r
= (Z/n)² /(n² /Z)
= Z3/n4
so acceleration, a ∝1/n⁴
now for m shell, n = 3
for l shell, n = 2
now initial acceleration/final acceleration = a ₁/a ₂
= (n² /n1)4 = (2/3)⁴ = 16/81
Therefore the ratio of initial to final acceleration is 16:81.
Question 4
1 / -0.25

The electron in a hydrogen atom makes a transition n₁ →n₂ whose n₁ and n₂ are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The frequency of orbital motion of the electron in the initial state is 1/27 of that in the final state. The possible values of n₁ and n₂ are

A
n₁ = 4, n₂ = 2

B
n₁ = 3, n₂ = 1

C
n₁ = 8, n₂ = 1

D
n₁ = 6, n₂ = 3

Solution

frequency ∝z² /n³
⇒f=kz² /n³
∴f1 =(1/27) f₂
⇒kz₁² /n₁³ =(1/27) (kz₂² /n₂³ ) (z₁ =z₂ =1)
⇒k/n₁³ =(1/27) (k/n₂³ )
⇒n₁ =3n₂
∴n₁ =3 and n₂ =1
Question 5
1 / -0.25

The radius of Bohr 's first orbit is a₀ . The electron in n^th orbit has a radius

A
na₀

B
a₀ /n

C
n² a₀

D
a₀ /n²

Solution

Radius of nth orbital
r_n = ϵ₀ n₂ h₂ / πmZe₂
Wherein
=r_n ∝n² /Z
= ϵ₀ h² /πme² =0.529
r=(n² /Z)a₀
For Z=1 r=n² a₀
Question 6
1 / -0.25

The ionisation potential of hydrogen atom is 13.6 volt. The energy required to remove an electron from the second orbit of hydrogen is :

A
3.4 eV

B
6.8 eV

C
13.6 eV

D
27.2 eV

Solution

Ionization energy-
E_ion =E_∞ -E_n
=13.6 (Z² /n² )
- wherein
Energy required to move an electron from ground state to
n=∞
Energy required to remove electron from 2nd bohr orbit is
△E=E₀ . [1/{2² -(1/∞)}]=E₀ /4=(13.6/4) eV
△E=3.4eV
Question 7
1 / -0.25

In a sample of hydrogen like atoms all of which are in ground state, a photon beam containing photons of various energies is passed. In absorption spectrum, five dark lines are observed. The number of bright lines in the emission spectrum will be (Assume that all transitions take place)

A
5

B
10

C
15

D
None of these

Solution
Question 8
1 / -0.25

When a hydrogen atom, initially at rest emits, a photon resulting in transition n = 5 → n = 1, its recoil speed is about

A
10^-4 m/s

B
2 ×10^-2 m/s

C
4.2 m/s

D
3.8 ×10^-2 m/s

Solution

Given,
n₁ =1
n₂ =5
m=1.6 ×10⁻²⁷ kg
Momentum, P=mv=h /λ
recoil speed, v= h /m λ. . . . . . . . .(1)
we know that,
1/λ=R[(1/n₁² )−(1/n₂² )]
1/λ=1.097 ×107[(1/1² )−(1/5² )]
λ=9.48 ×10⁻⁸ m
From equation (1),
recoil speed, v=6.67 ×10⁻³⁴ /1.6 ×10⁻²⁷ ×9.48 ×10⁻⁸ =4.2m/s
The correct option is C.
Question 9
1 / -0.25

Consider the spectral line resulting from the transition n = 2 → n = 1 in the atoms and ions given below. The shortest wavelength is produced by :

A
Hydrogen atom

B
Deuterium atom

C
Singly ionized helium

D
Doubly ionized lithium

Solution

Wavelength is calculated as
1/λ=Rz² (1/n_f² =1/n_i² )
∴λ∝1/z²
For hydrogen atom, z=1
For Deuterium atom, z=2
For singly ionized helium, z=2
For doubly ionized lithium, z=3
Thus, z is maximum for lithium and so λis minimum.
Hence the correct answer is option D.
Question 10
1 / -0.25

In an atom, two electrons move around the nucleus in circular orbits of radii R and 4R. The ratio of the time taken by them to complete one revolution is : (neglect electric interaction)

A
1 : 4

B
4 : 1

C
1 : 8

D
8 : 1

Solution

Time period, T ∝(R)3/2
∴T1/T2 = (R1/R2 )3/2= (1/4 )3/2=1/8
Thus, the ratio of time period is 1:8

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Dual Nature of Matter and Radiation Test - 9

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Dual Nature of Matter and Radiation Test - 9

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Question 1

6

7

8

5

Solution

Question 2

1

2

3

4

Solution

Question 3

9 : 4

81 : 16

4 : 9

16 : 81

Solution

Question 4

n1 = 4, n2 = 2

n1 = 3, n2 = 1

n1 = 8, n2 = 1

n1 = 6, n2 = 3

Solution

Question 5

na0

a0 /n

n2 a0

a0 /n2

Solution

Question 6

3.4 eV

6.8 eV

13.6 eV

27.2 eV

Solution

Question 7

5

10

15

None of these

Solution

Question 8

10-4 m/s

2 ×10-2 m/s

4.2 m/s

3.8 ×10-2 m/s

Solution

Question 9

Hydrogen atom

Deuterium atom

Singly ionized helium

Doubly ionized lithium

Solution

Question 10

1 : 4

4 : 1

1 : 8

8 : 1

Solution

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n₁ = 4, n₂ = 2

n₁ = 3, n₂ = 1

n₁ = 8, n₂ = 1

n₁ = 6, n₂ = 3

na₀

a₀ /n

n² a₀

a₀ /n²

10^-4 m/s

2 ×10^-2 m/s

3.8 ×10^-2 m/s