Question

An X-ray pulse of wavelength 4.9 $\mathring{A}$ is sent through a section of Wilson cloud chamber containing a super saturated gas, and tracks of photoelectron ejected from the gaseous atoms are observed. Two groups of tracks of lengths 1.40 cm and 2.02 cm are noted. If the range-energy relation for cloud chamber is given by R = $\alpha$ E with $\alpha =1cmke{V}^{-1}$, obtain the binding energies of the two levels from which electroris are emitted. Given h = 6.63 $\times 10^{-34}Js,e=1.6\times 10^{-19}$ J.

• A. 0.52 ke V
• B. 0.75 eV
• C. 0.52 eV
• D. 0.75 keV

Answer 1

Answer: (A)

Binding energy, W = $\frac{hc}{\lambda }-E$
where E = $\frac{R}{\alpha }$
Here, $\lambda =4.9\times 10^{-10}m,R_1=1.4cm,R_2=2.02$ cm,
$\alpha =1cmke{V}^{-1}$
For cloud chamber, the range-energy reiation is
R = $\alpha E$ or $E=\frac{R}{\alpha }$
$\therefore E_1=\frac{R_1}{\alpha }=\frac{140}{1}=1.40keV$
and $E_1=\frac{R_2}{\alpha }=\frac{2.02}{1}=2.02$ ke V
Energy of the incident photon
hv = $\frac{hc}{\lambda }=\frac{6.63\times 10^{-34}\times 3\times 10^8}{4.9\times 10^{-10}}J$
= $\frac{6.63\times 3\times 10^{-16}}{4.9\times 1.6\times 10^{-16}}$ ke V
= 2.54 ke V
From Einstein's photoelectric equation
hv = W + E
$\therefore$ Binding energy,
W = hv - E
or $W_1=2.54-1.40=1.14keV$
and $W_2=2.54-2.02$
= 0.52 keV