In a Frank-Hertz experıment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV . The minimum wavelength of photons emitted by mercury atoms is close to
Select the correct option:
A
1700 nm
B
2020 nm
C
250 nm
D
220 nm
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \text { Energy lost by electron }=5.6-0.7=4.9 \mathrm{eV} \\ & \frac{\mathrm{hc}}{\lambda_{\min }}=4.9 \quad \Rightarrow \lambda_{\min }=\frac{1240}{4.9}=250 \mathrm{~nm}\end{aligned}$
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