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 $

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 12. atoms→12. atoms — all question types→Physics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

The idea of secondary wavelets for the propagation of a wave was first given by

Electric potential of earth is taken to be zero because earth is a good

If the monochromatic source in Young's double slit experiment is replaced by white light, then

What is the effective capacitance between A and B in the following figure

Consider two equivalent, triangular hollow prisms $A$ and $B$ made of thin glass plates and arranged with negligible spacing as shown in the figure. A beam of white light is incident on prism $A$ from the left. Given that, the refractive index of water is inversely related to temperature, the beam to the right of prism $B$ would not appear white, if

In the figure, potential difference between $A$ and $B$ is......$V$

Lenz's law gives

Light waves can be polarised because they:

A ray of light incidents on a plane mirror at an angle of 30°. The deviation produced in the ray is

Consider the circuits shown in the figure. Both the circuits are taking same current from battery but current through $R$ in the second circuit is $\frac{1}{{10}}$$^{th}$ of current through $R$ in the first circuit. If $R$ is $11$ $\Omega$, the value of $ R_1$ ................ $\Omega$