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 = 1 \, cm \, ke 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 = 1.4 \, cm, \, R_2 = 2.02$ cm,
$ \alpha = 1 \, cm \, ke \, 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} = 1.40\, k e \, V $
and $ E_1 = \frac{ R_2}{ \alpha } = \frac{ 2.02 }{ 1} = 2.02 $ ke V
Energy of the incident photon
hv = $ \frac{ h c }{ \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 = 1.14 \,k e \,V $
and $ W_2 = 2.54 - 2.02 $
= 0.52 keV