Question

A $600W$ mercury lamp emits monochromatic radiation of wavelength $331.3nm$. How many photons are emitted from the lamp per second?
($h = 6.626 \times {10^{ - 34}}J - s$; velocity of light$ = 3 \times {10^8}m{s^{ - 1}}$)
A) $1 \times {10^{19}}$
B) $1 \times {10^{20}}$
C) $1 \times {10^{21}}$
D) $1 \times {10^{23}}$

Answer 1

Hint:Light radiations exhibit both wave and particle nature and this phenomenon of radiations is known as dual nature of light. In particle form radiations are made up of small packets of energy which we referred to as photon and the energy of these photon depends upon radiation wavelength
Formula Used:
$n = \dfrac{{P\lambda }}{{hc}}$
Here $n$is the number of photons emitted per second
$P$ is the power of lamp in $W$
$\lambda $is the wavelength of radiation in$m$
$c$ is velocity of light$ = 3 \times {10^8}m{s^{ - 1}}$)
$h = 6.626 \times {10^{ - 34}}J - s$ (Planck constant)

Complete step by step answer:
As given in the question
$P = 600W$
Wavelength $\left( \lambda \right) = $$331.3nm$
As we know $1nm = {10^{ - 9}}m$
Then Wavelength $\left( \lambda \right)$ in meters $ = 331.3 \times {10^{ - 9}}m$
$h = 6.626 \times {10^{ - 34}}J - s$ (Planck constant)
$c$ $ = 3 \times {10^8}m{s^{ - 1}}$


Using all the value in the given formula
$n = \dfrac{{P\lambda }}{{hc}}$
$n = \dfrac{{600 \times 331.3 \times {{10}^{ - 9}}}}{{3 \times {{10}^8} \times 6.626 \times {{10}^{ - 34}}}}$
$n = 1 \times {10^{21}}$
Hence option ‘C’ is the correct solution for the given question.

Additional Information: A mercury-vapor lamp is a gas-discharge lamp which uses an electric arc through vaporized mercury to produce light. The arc discharge is generally confined to a small fused quartz arc tube mounted within a larger borosilicate glass bulb. The outer bulb should be clear or coated with a phosphor; in most of the cases, the outer bulb is used for thermal insulation, and provides protection from the ultraviolet radiation that the light produces, and a convenient mounting for the fused quartz arc tube.
They work at an internal pressure at one atmosphere and need some special fixtures, as well as an electrical ballast.

Note:
The above used formula in the question can be updated if energy of radiation is given instead of power of lamp as follow
$n = \dfrac{{E \times \lambda }}{{h \times c \times t}}$
Where
 $n$ is the number of photons emitted per second
$E$ is the energy of the radiation in $Joules$
$\lambda $ is the wavelength of radiation in $m$
$c$ is velocity of light$ = 3 \times {10^8}m{s^{ - 1}}$)
$h = 6.626 \times {10^{ - 34}}J - s$ (Planck constant)
$t$ is the time for which radiation was emitted in $\sec $