An electron passing through a potential difference of 4.9 V colli… - Vidyadip

MCQ

An electron passing through a potential difference of $4.9 V$ collides with a mercury atom and transfers it to the first excited state. What is the wavelength of a photon corresponding to the transition of the mercury atom to its normal state.......$\mathop A\limits^o $

A
$2050 $
B
$2240 $
✓
$2525 $
D
$2935 $

✓

Answer

Correct option: C.

$2525 $

c
(c) $\frac{{hc}}{\lambda } = E = eV$

$ \Rightarrow \lambda = \frac{{hc}}{{eV}} - \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.6 \times {{10}^{ - 19}} \times 4.9}} = 2525\mathop A\limits^ \circ $

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