JEE Main Atom PYQ

46. (JEE Main 2018 (Offline))

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is p_d; while for its similar collision with carbon nucleus at rest, fractional loss of energy is p_c. The values of p_d and p_c are respectively :

A. (0, 1)

B. (0.89, 0.28)

C. (0.28, 0.89)

D. (0, 0)

The correct answer is option (C)

JEE Main 2018 (Offline) Physics - Atoms and Nuclei Question 174 English Explanation 1

Applying conservation of momentum :

mv + 0 = mv₁ + 2mv₂

$\Rightarrow$ v = v₁ + 2v₂ . . . . . (1)

As collision is elastic,

So, coefficient of restitution, e = 1

$∴$ e = 1 = $\frac{v e l o c i t y o f s e p a r a t i o n}{V e l o c i t y o f a p p r o a c h}$

$\Rightarrow$ 1 = $\frac{v_{2} - v_{1}}{v - 0}$

$\Rightarrow$ v = v₂ $-$ v₁ . . . . .(2)

Add (1) and (2),

2v = 3v₂

$\Rightarrow$ v₂ = $\frac{2 v}{3}$

put value of v₂ in equation (1),

v₁ = v $-$ 2v₂

= v $-$ $\frac{4 v}{3}$

= $-$ $\frac{v}{3}$

$∴$ Fractional loss of energy of neutron.

P_d = $\frac{k_{i} - k_{f}}{k_{i}}$

= $\frac{\frac{1}{2} m v^{2} - \frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v^{2}}$

= $\frac{v^{2} - \frac{v^{2}}{9}}{v^{2}}$

= $\frac{8}{9}$

= 0.89

JEE Main 2018 (Offline) Physics - Atoms and Nuclei Question 174 English Explanation 2

Applying momentum of conservation,

mv + 0 = mv₁ + 12mv₂

$\Rightarrow$ v = v₁ + 12v₂ . . . . . (3)

Here also e = 1

$∴$ e = 1 = $\frac{v_{2} - v_{1}}{v - 0}$

$\Rightarrow$ v = v₂ $-$ v₁ . . . . . . (4)

adding (3) and (4), we get

2v = 13v₂

$\Rightarrow$ v₂ = $\frac{2 v}{13}$

put this v₂ in equation (3), we get

v₁ = v $-$ 12 $\times$ $\frac{2 v}{13}$

= $-$ $\frac{11 v}{13}$

$∴$ Frictional loss

p_c = $\frac{\frac{1}{2} m v^{2} - \frac{1}{2} m {(\frac{11}{13} v)}^{2}}{\frac{1}{2} m v^{2}}$

= $\frac{48}{169}$

= 0.28

47. (JEE Main 2017 (Offline))

Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r =

λ_{1}

/

λ_{2}

, is given by:

JEE Main 2017 (Offline) Physics - Atoms and Nuclei Question 173 English

A. r = 1/3

B. r = 4/3

C. r = 2/3

D. r = 3/4

The correct answer is option (C)

From energy level diagram, using $Δ E = \frac{h c}{λ}$

For wavelength $λ_{1}$ , $Δ E =$ –E – (–2E) = $\frac{h c}{λ_{1}}$

For wavelength $λ_{2}$ , $Δ E =$ –E – (– $\frac{4 E}{3}$ ) = $\frac{h c}{λ_{2}}$

$∴$ r = $\frac{λ_{1}}{λ_{2}} = \frac{(\frac{4 E}{3} - E)}{(2 E - E)}$ = $\frac{1}{3}$

48. (JEE Main 2016 (Online) 9th April Morning Slot)

A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :

A. 2

B. 3

C. 4

D. 5

The correct answer is option (C)

Energy released when hydrogen atom makes transition from n = 2 to n = 1 is,

E₁ = 13.6 $\times$ $(\frac{1}{1^{2}} - \frac{1}{z^{2}})$

= $\frac{3}{4} \times 13.6$ eV

Energy required to remove a electron from n^th excited state of doubly ionized lithium,

E₂ = $\frac{13.6 z^{2}}{n^{2}}$ = $\frac{13.6 \times 3^{2}}{n^{2}}$ eV

This energy is provided by the photon when it strike with the lithium atom.

$∴$    E₁ $\geq$ E₂

$\Rightarrow$     $\frac{3}{4} \times 13.6 \geq \frac{13.6 \times 9}{n^{2}}$

$\Rightarrow$    n²    $\geq$   3 $\times$ 4

$\Rightarrow$   n    $\geq$ $\sqrt{12}$

$\Rightarrow$     n $\geq$ 3.5

$∴$   Least possible excited state = 4

49. (JEE Main 2014 (Offline))

Hydrogen

(_{1} H^{1})

, Deuterium

(_{1} H^{2})

, singly ionised Helium

{(_{2} H e^{4})}^{+}

and doubly ionised lithium

{(_{3} L i^{6})}^{+ +}

all have one electron around the nucleus. Consider an electron transition from

n = 2

to

n = 1.

If the wavelengths of emitted radiation are

λ_{1}, λ_{2}, λ_{3}

and

λ_{4}

respectively then approximately which one of the following is correct?

A. $4 λ_{1} = 2 λ_{2} = 2 λ_{3} = λ_{4}$

B. $λ_{1} = 2 λ_{2} = 2 λ_{3} = λ_{4}$

C. $λ_{1} = λ_{2} = 4 λ_{3} = 9 λ_{4}$

D. > $λ_{1} = 2 λ_{2} = 3 λ_{3} = 4 λ_{4}$

The correct answer is option (C)

Wave number $\frac{1}{λ} = R Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n^{2}}]$

$\Rightarrow λ \propto \frac{1}{Z^{2}}$

By question $n = 1$ and $n_{1} = 2$

Then, $λ_{1} = λ_{2} = 4 λ_{3} = 9 λ_{4}$

50. (JEE Main 2014 (Offline))

The radiation corresponding to

3 \to 2

transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field

3 \times 10^{- 4} T .

If the radius of the larger circular path followed by these electrons is

10.0

m m

, the work function of the metal is close to:

A. $1.8$ $e V$

B. $1.1$ $e V$

C. $0.8$ $e V$

D. $1.6$ $e V$

The correct answer is option (B)

Radius of circular path followed by electron is given by,

$r = \frac{m υ}{q B} = \frac{\sqrt{2 m e V}}{e B} = \frac{1}{B} \sqrt{\frac{2 m}{e} V}$

$\Rightarrow V = \frac{B^{2} r^{2} e}{2 m} = 0.8 V$

For transition between $3$ to $2.$

$E = 13.6 (\frac{1}{4} - \frac{1}{9})$

$= \frac{13.6 \times 5}{36} = 1.88 e V$

Work function $= 1.88 e V - 0.8 e V$

$= 1.08 e V \approx 1.1 e V$