Q 11 :

The radius of 2nd orbit of He+ of Bohr's model is r1 and that of fourth orbit of Be3+ is represented as r2. Now the ratio r2r1 is x:1. The value of x is _________    [2023]



(2)

r∝n2z

rHe+rBe3+=22×42×4×4=12



Q 12 :

In hydrogen atom spectrum, (R→Rydberg's constant)

A.   the maximum wavelength of the radiation of Lyman series is 43R
B.   the Balmer series lies in the visible region of the spectrum
C.   the minimum wavelength of the radiation of Paschen series is 9R
D.   the minimum wavelength of Lyman series is 54R

Choose the correct answer from the options given below:                             [2026]

  • A, B Only

     

  • A, B and C Only

     

  • A, B and D Only

     

  • B, D Only

     

(2)

1λ=R(1-14)

λ=43R

1λ'=R(19)

λ'=9R



Q 13 :

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ______ m 
(Atomic number of gold = 79 and 14πε0=9×109 in SI units)    [2026]

  • 3.85×10-16

     

  • 2.95×10-14

     

  • 3.85×10-14

     

  • 2.95×10-16

     

(2)

Energy conservation

Ki+Ui=Kf+Uf

7.7×106×1.6×10-19+0

=0+9×109(1.6×10-19)(79×1.6×10-19)r

r=2.95×10-14



Q 14 :

Two electrons are moving in orbits of two hydrogen-like atoms with speeds 3×105 m/s and 2.5×105 m/s respectively. If the radii of these orbits are nearly same, then the possible order of energy states are _______ respectively.             [2026]

  • 8 and 10

     

  • 6 and 5

     

  • 10 and 12

     

  • 9 and 8

     

(2)

V∝Zn

r∝n2Z

Thus, r∝nV

Radii are same then

n1V1=n2V2

n1n2=3×1052.5×105=65

Possible order is 6 and 5



Q 15 :

7.9 MeV α-particle scatters from a target material of atomic number 79. From the given data, the estimated diameter of nuclei of the target material is (approximately) _____ m.


[14πε0=9×109 Nm2/C2 and electron charge =1.6×10-19 C]                      [2026]

  • 1.44×10-13

     

  • 2.88×10-14

     

  • 1.69×10-12

     

  • 5.76×10-14

     

(4)

By mechanical energy conservation

(ME)i=(ME)f

PEi+KEi=PEf+KEf

0+7.9×106×1.6×10-19=k(2e)(Ze)r+0

r=9×109×2×(1.6×10-19)2×797.9×106×1.6×10-19=2.88×10-14 m

For diameter ⇒D=2r=5.76×10-14 m



Q 16 :

The energy of an electron in an orbit of the Bohr’s atom is -0.04E0 eV where E0 is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck’s constant, then 2πLh=_____.   [2026]

  • 5

     

  • 2

     

  • 6

     

  • 4

     

(1)

Angular momentum L=nh2π

n=2πLh

Energy E=-13.6n2Z2

E⇒-E0n2=-0.04E0

n2=25,  n=5



Q 17 :

According to Bohr model, magnetic field at centre (at the nucleus) of a hydrogen atom due to motion of electron in the nth orbit is proportional to

  • 1n3

     

  • 1n5

     

  • n5

     

  • n3

     

(2)

B=μ0i2r; magnetic field at centre of hydrogen atom i.e. at nucleus.

i=eT=ef∝Z2n3

r∝n2Z

B∝ir∝Z3n5

B∝1n5



Q 18 :

The minimum energy of an electron necessary for ionization of an atom of hydrogen, is equal to W0. Suppose that a hydrogen atom and helium atom both in ground state collide inelastically so that hydrogen atom is ionized. If the hydrogen was at rest and helium was moving with at least energy of xW0, find the value of x.



(5)

Possible loss=124m5V2=W0

KEHe=124mV2=5W0



Q 19 :

Hydrogen gas absorbs radiations of wavelength λ0 and consequently emit radiations of 6 different wavelengths of which three wavelengths are shorter than λ0. Choose the CORRECT statement(s).

  • The final energy state of the atom is n=4

     

  • The initial state of the atom may be n=2

     

  • There are two transitions in Balmer series

     

  • There are three transitions belonging to Lyman series

     

Select one or more options

(1,2,3,4)

Since only 6 different wavelengths are excited, therefore highest excited state is n=4.

Two wavelengths are shorter than λ0. Initially atoms were in excited state n=2.

Corresponding transitions are 4 → 3, 4 → 2, 4 → 1, 3 → 2, 3 → 1, 2 → 1.

 



Q 20 :

An electron in a hydrogen atom makes a transition n1→n2, where n1 and n2 are the principal quantum numbers of the two states. Assume Bohr model to be valid.

  Column-I   Column-II
A) The electron emits an energy of 2.55 eV p) n1=2, n2=1
B) Time period of the electron in the initial state is eight times that in the final state q) n1=4, n2=2
C) Speed of electron become two times r) n1=5, n2=3
D) Radius of orbit of electron becomes 4.77 A0 s) n1=6, n2=3
    t) n1=8, n2=4

 

  • A→p,   B→p,t,   C→p,q,t,   D→r

     

  • A→p,   B→p,t,   C→p,q,s,t,   D→r

     

  • A→q,   B→p,q,s,t,   C→p,q,s,t,   D→r,s

     

  • A→r,   B→p,   C→p,q,s,t,   D→q

     

(3)

En=-13.6Z2n2eV,  rn=0.529n2ZÅ,  Vn=2.2×106Znm/s,  T=2πrnvn.



Q 21 :

Angular momentum of an electron in a hydrogen atom is 3hπ, then the energy of the electron is ______ eV.             [2026]

  • − 1.51

     

  • − 0.85

     

  • − 0.38

     

  • − 0.28

     

(3)

L=nh2π=3hπ⇒n=6

E=-13.6n2eV=-13.636eV≈-0.38 eV



Q 22 :

In Rutherford’s alpha-particle scattering experiment, only a few alpha particles rebound back because

(A) The size of gold nucleus is very small as compared to the size of gold atom.

(B) Alpha particle and gold nucleus have equal charge.

(C) The impact parameter is minimum for a few alpha particles.

(D) A few alpha particles have very high kinetic energy.

(E) Only a few alpha particles undergo head-on collision with the nuclei.

Choose the correct answer from the options given below:                               [2026]

  • A, B Only

     

  • B, E Only

     

  • C, D Only

     

  • A, C, E Only

     

(4)

(A) size of nucleus is small

(C) impact parameter is zero only for few α-particles

(E) only few α-particles can rebound by 180° by head-on collision.



Q 23 :

In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit (f). The ratio of the radius of the orbits is given by ri:rf=16:4. The wavelength of photon emitted due to this transition is ______ nm. (Given Rydberg constant = 1.0973×107/m)              [2026]
 

  • 121

     

  • 242

     

  • 486

     

  • 974

     

(3)

Given:  rirf=164=ni2nf2

ninf=42

1λ=R(1nf2-1ni2)=1.0973×107(1(2)2-1(4)2)

λ=4.86×10-7 m≈486 nm



Q 24 :

Using Bohr’s model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the 2nd and 4th orbits of hydrogen atom _________.          [2026]



(64)

The magnetic field due to circular loop,  B=μ0I2πr

Current due to orbiting electron,  I=ev2πr

⇒B∝vr2

But,  v∝Zn  and  r∝n2Z

⇒B∝(Zn)(n2Z)2=Z3n5

B1B2=n25n15=4525=64