Question

Photoelectrons are emitted when 400 nm radiation is incident on a surface of work function 1.9 eV. These photoelectrons pass through a region containing $\alpha -particles$. A maximum energy electron combines with an $\alpha -particles$ to form a $H{e}^{+}$ ions, emitting a single photon in this process. $H{e}^{+}$ ions thus formed are in their fourth excited state. Find out the number of photon whose energies, lying in the 2 to 3 eV range, that are likely to be emitted during and after the combination. [Take $h=4.14\times {10}^{-15}eVs$]

Answer 1

From Einstein's photoelectric equation, maximum kinetic energy of emitted electrons
$E_k=\frac{hc}{\lambda }-W$
Given $h=4.14\times {10}^{-15}eV-s$
$=4.14\times {10}^{-15}\times 1.6\times {10}^{-19}J-s$
$=6.624\times {10}^{-34}J-s$
$\therefore E=\frac{hc}{\lambda }=\frac{6.624\times {10}^{-34}\times 3\times {10}^8}{400\times {10}^{-9}}$
$=4.968\times {10}^{-19}J$
$=\frac{4.968\times {10}^{-19}}{1.6\times {10}^{-19}}=3.1eV$
$E_k=3.1eV-1.9eV$
$=1.2eV$
$e+He\to H{e}^{+}+Photon$
Energy of He atom in their fourth $(n=5)$ excited state
$E_n=-\frac{Z^{2}Rhc}{n^2}=-\frac{13.6Z^2}{n^2}eV$
$=-\frac{(2{)}^2\times 13.6}{(5{)}^2}=-2.175eV$
(for $H{e}^{+}$ ion $Z=2)$
From conservation of energy,
$1.2eV+0=-2.176eV+E_{\gamma }$
Energy of photon during combination,
$E_{\gamma }=1.2+2.176=3.376eV$
Energy of Helium ion,
$E_n=-frac{Z}^{2}Rhc{n}^2$
$=-\frac{4\times 13.6}{n^2}$
$=-\frac{54.4}{n^2}eV,n=1,2,3,....$
$=-54.4eV,-13.6eV,-6.04eV,-3.4eV,-2.176ev,-1.51eV$
Difference of energies lying between 2 and 4 eV is
$-3.4+6.04=2.64eV$
$-2.176+6.04=3.86eV$
Energies of photons emitted are 2.64 eV and 3.86 eV.