Question

The vapours of Hg absorb some electrons accelerated
by a potential difference of 4.5 volt as a result of which
light is emitted. If the full energy of single incident e is
supposed to be converted into light emitted by single
Hg atom, find the wave number of the light.
(A) 3.63 x 10 m
(B) 36.3 x 10m
(C)3.63 x 10 m
(D) 3.63 x 107m'​

Answer 1

Given:

The vapours of Hg absorb some electrons accelerated by a potential difference of 4.5 volt as a result of which light is emitted.

To find:

Find the wave number of the light.

Solution:

From given, we have,

A potential difference of 4.5 volt

⇒ E = 4.5 V

we use the formula,

E = hc/λ

4.5 = hc/λ

we know that,

The wavenumber is reciprocal of the wavelength.

So, ν = 1/λ

⇒ 4.5 = hcν

⇒ ν = 4.5/hc

⇒ ν = 4.5 / (6.63 × 10-34 × 3 × 10^8)

∴ ν = 4.5 eV / 1242 eV-nm

∴ ν = 3.63 × 10^6 m-1

Therefore, the wave number is 3.63 × 10^6 m-1

Answer 2

Answer:

3.63x10^6 m^-1

is the correct answer