An $X$-ray pulse of wavelength $4.9\ \mathring{\mathrm{A}}$ 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 $\alpha = 1cm/keV,$ obtain the binding energies of the two levels from which electrons are emitted. Given $h = 6.63 imes 10^{- 34}J - s,e = 1.6 imes 10^{- 19}J$

Question

An $X$-ray pulse of wavelength $4.9\ \mathring{\mathrm{A}}$ 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 $\alpha = 1cm/keV,$ obtain the binding energies of the two levels from which electrons are emitted. Given $h = 6.63 	imes 10^{- 34}J - s,e = 1.6 	imes 10^{- 19}J$

Dr. Abubacker · Accepted Answer

(a) Binding energy, $W = \frac{hc}{\lambda} - E;$
Where $E = R/\alpha;$ Here$,\ \lambda = 4.9 	imes 10^{- 10}m$
$R_{1} = 1.4cm,\ R_{2} = 2.02cm;\ \alpha = 1cm/keV$
For cloud chamber, the range-energy relation is
$R = \alpha E\ $or$\ E = R/\alpha$
$	herefore E_{1} = \frac{R_{1}}{\alpha} = \frac{1.40}{1} = 1.40keV$
$and\ E_{2} = \frac{R_{2}}{\alpha} = \frac{2.02}{1} = 2.02keV$
Energy of the incident photon
$hv = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{- 34} 	imes 3 	imes 10^{8}}{4.9 	imes 10^{- 10}}J = 2.54\ keV$
From Einstein’s Photoelectric equation
$hv = W + E$
$	herefore$ Binding energy$,\ W = hv - E$
$\Rightarrow W_{1} = 2.54 - 1.40 = 1.14\ keV$
and$\ W_{2} = 2.54 - 2.02 = 0.52\ keV$

Explanation

Reviewed by Dr. Abubacker

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