SATHEE: Modern Physics 2 Question 34

Modern Physics 2 Question 34

33. Photoelectrons are emitted when $400 n m$ radiation is incident on a surface of work function $1.9 e V$ . These photoelectrons pass through a region containing $α$ -particles. A maximum energy electron combines with an $α$ -particle to form a $H e^{+}$ ion, emitting a single photon in this process. $H e^{+}$ ions thus formed are in their fourth excited state. Find the energies in $e V$ of the photons lying in the 2 to $4 e V$ range, that are likely to be emitted during and after the combination. (1999, 5M) [Take $h = 4.14 \times 10^{- 15} e V$ -s]

Show Answer

Answer:

Correct Answer: 33. During combination $3.4 e V$ . After combination $3.84 e V, 2.64 e V$ 34. (a) $10$ (b) 285.1 (d) $111 s$

Solution:

Given work function, $W = 1.9 e V$

Wavelength of incident light, $λ = 400 n m$

$∴$ Energy of incident light, $E = \frac{h c}{λ} = 3.1 e V$

(Substituting the values of $h, c$ and $λ$ )

Therefore, maximum kinetic energy of photoelectrons

$K_{max} = E - W = (3.1 - 1.9) = 1.2 e V$

Now the situation is as shown below :

Energy of electron in $4^{th}$ excited state of $H e^{+} (n = 5)$ will be

$E_{5} = - 13.6 \frac{Z^{2}}{n^{2}} e V \Rightarrow E_{5} = - (13.6) \frac{(2)^{2}}{(5)^{2}} = - 2.2 e V$

Therefore, energy released during the combination

$= 1.2 - (- 2.1) = 3.4 e V$

Similarly, energies in other energy states of $H e^{+}$ will be

$\begin{aligned} E_{4} = - 13.6 \frac{(2)^{2}}{(4)^{2}} = - 3.4 e V \\ E_{3} = - 13.6 \frac{(2)^{2}}{(3)^{2}} = - 6.04 e V \\ E_{2} = - 13.6 \frac{(2)^{2}}{2^{2}} = - 13.6 e V \end{aligned}$

The possible transitions are

$\begin{aligned} Δ E_{5 \to 4} & = E_{5} - E_{4} = 1.2 e V < 2 e V \\ Δ E_{5 \to 3} & = E_{5} - E_{3} = 3.84 e V \\ Δ E_{5 \to 2} & = E_{5} - E_{2} = 11.4 e V > 4 e V \\ Δ E_{4 \to 3} & = E_{4} - E_{3} = 2.64 e V \\ Δ E_{4 \to 2} & = E_{4} - E_{2} = 10.2 e V > 4 e V \end{aligned}$