SATHEE: Modern Physics 7 Question 48

Modern Physics 7 Question 48

54. A neutron of kinetic energy $65 e V$ collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of $90^{\circ}$ with respect of its original direction. $(1993, 9 + 1 M)$

(a) Find the allowed values of the energy of the neutron and that of the atom after the collision.

(b) If the atom gets de-excited subsequently by emitting radiation, find the frequencies of the emitted radiation.

[Given : Mass of He atom $= 4 \times ($ mass of neutrons $)$

Ionization energy of $H$ atom $= 13.6 e V]$

Show Answer

Answer:

Correct Answer: 54. (a) $2.87 \times 10^{13} s^{- 1} m^{- 2}, 2.07 \times 10^{- 10} A$

(b) $2.06 \times 10^{13} s^{- 1} m^{- 2}, 1.483 \times 10^{- 10} A$ (c) $1.06 V$ in both cases

Solution:

(a) Let $K_{1}$ and $K_{2}$ be the kinetic energies of neutron and helium atom after collision and $Δ E$ be the excitation energy.

From conservation of linear momentum along $x$ -direction.

$\begin{aligned} p_{i} & = p_{f} \\ \Rightarrow \sqrt{2 K m} & = \sqrt{2 (4 m) K_{2}} \cos θ \end{aligned}$

Similarly, applying conservation of linear momentum in $y$ -direction, we have

$\sqrt{2 K_{1} m} = \sqrt{2 (4 m) K_{2}} \sin θ$

Squaring and adding Eqs. (i) and (ii), we get

$\begin{aligned} K + K_{1} & = 4 K_{2} \\ or 4 K_{2} - K_{1} & = K = 65 e V \end{aligned}$

Now, during collision, electron can be excited to any higher energy state. Applying conservation of energy, we get $K = K_{1} + K_{2} + Δ E$

or $65 = K_{1} + K_{2} + Δ E$

$Δ E$ can have the following values,

$Δ E_{1} = - 13.6 - (- 54.4) e V = 40.8 e V$

Substituting in (v), we get

$K_{1} + K_{2} = 24.2 e V$

Solving (iv) and (vi), we get

$\begin{array}{ll} K_{1} = 6.36 e V \\ and & K_{2} = 17.84 e V \end{array}$

Similarly, when we put $Δ E = Δ E_{2}$

$\begin{aligned} = - 6.04 - (- 54.4) e V \\ = 48.36 e V \end{aligned}$

Put in Eq. (v), we get

$K_{1} + K_{2} = 16.64 e V$

Solving Eqs. (iv) and (vii), we get

$K_{1} = 0.312 e V and K_{2} = 16.328 e V$

Similarly, when we put

$Δ E = Δ E_{3} = - 3.4 - (- 54.4) = 51 e V$

Put in Eq. (v), we get

$K_{1} + K_{2} = 14 e V$

Now, solving Eqs. (iv) and (viii), we get

$K_{1} = - 1.8 e V and K_{2} = 15.8 e V$

But since the kinetic energy cannot have the negative values, the electron will not jump to third excited state or $n = 4$ .

Therefore, the allowed values of $K_{1}$ ( $K E$ of neutron) are $6.36 e V$ and $0.312 e V$ and of $K_{2}$ (KE of the atom) are 17.84 $e V$ and $16.328 e V$ and the electron can jump upto second excited state only $(n = 3)$ .

(b) Possible emission lines are only three as shown in figure. The corresponding frequencies are

$v_{1} = \frac{(E_{3} - E_{2})}{h}$

$= \frac{- 6.04 - (- 13.6) \times 1.6 \times 10^{- 19}}{6.63 \times 10^{- 34}}$

$\begin{aligned} = 1.82 \times 10^{15} H z \\ v_{2} & = \frac{E_{3} - E_{1}}{h} = \frac{- 6.04 - (- 54.4) \times 1.6 \times 10^{- 19}}{6.63 \times 10^{- 34}} \\ = 11.67 \times 10^{15} H z \\ and v_{3} & = \frac{E_{2} - E_{1}}{h} \\ = \frac{- 13.6 - (- 54.4) \times 1.6 \times 10^{- 19}}{6.63 \times 10^{- 34}} \\ = 9.84 \times 10^{15} H z \end{aligned}$