Question

Photoelectrons are emitted when $400nm$ radiation is incident on a surface of work function $1.9eV$. These photoelectrons pass through a region containing $\alpha -particles$. A maximum energy electron combines with an $\alpha -particle$ 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 eV of the photons, lying in the $2$ to $4eV$ range, that are likely to be emitted during and after the combination.

• A. 3.376 eV, 3.94 eV, 2.64 eV
  
• B. 4.376 eV, 3.94 eV, 2.64 eV
  
• C. 3.376 eV, 5.94 eV, 2.64 eV
  
• D. 3.376 eV, 6.94 eV, 2.64 eV
  

Answer 1

The correct option is D 3.376 eV, 6.94 eV, 2.64 eV

$E_1=hc/\lambda =(4.14\times {10}^{-15}eVs)(3\times {10}^{8}m/s)/(4000\times {10}^{-9}m)=3.1eV$

The maximum kinetic energy of the electron is $E_{max}=E_1-W=3.1eV-1.9eV=1.2eV$

It is given that

(Emitted electrons of maximum energy)$+{}_{2}H{e}^{2+}\to H{e}^{+}$ in $4$ the excited state+photon

The fourth excited state implies that the electron enter tje $n=5$ state in this state its energy is

$E_5=-\left(13.6eV\right)Z^2/n^2=-\left(13.6eV\right)(2{)}^2/5^2$

$=-2.18eV$

$E=E_{max}+(-E_5)=1.2eV=2.4eV$

Note: After the recombination reaction, the electron may undergoes transition from a higher level to a lower level there by emitting photons

This energy of the emitted photon in teh above combination reaction is

The energies the electronic levels of ${He}^{+}$ are

$E_4=(-13.6eV)\left(2^2\right)/4^2=3.4eV$

$E_3=(-13.6eV)\left(2^2\right)/3^2=-6.04eV$

$E_2=(-13.6eV)\left(2^2\right)/2^2=-13.6eV$

Now, possible transitions are $\to$

$n=5\to n=4$

$\Delta E=E_5-E_4=[-2.18-(-3.4)]eV=12.8eV$

$n=5\to n=3$

$\Delta E=E_5-E_3=[-2.18-(-6.04)]eV=3.84eV$

$n=5\to n=2$

$\Delta E=E_5-E_2=[-2.18-(-13.6)]eV=11.4eV$

$n=4\to n=3$

$\Delta E=E_4-E_3=[-3.4-(-6.04)]eV=2.64eV$