# Neon gas is generally used in the sign boards.

Question.

Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate

(a) the frequency of emission,

(b) distance traveled by this radiation in 30 s

(c) energy of quantum and

(d) number of quanta present if it produces 2 J of energy.

Solution:

Wavelength of radiation emitted $=616 \mathrm{~nm}=616 \times 10^{-9} \mathrm{~m}$ (Given)

(a) Frequency of emission $(v)$

$v=\frac{c}{\lambda}$

Where, $c=$ velocity of

radiation $\lambda=$ wavelength of

Substituting the values in the given expression of $(v):$

$v=\frac{3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}}{616 \times 10^{-9} \mathrm{~m}}$

$=4.87 \times 10^{8} \times 10^{9} \times 10^{-3} \mathrm{~s}^{-1} \mathrm{~V}$

$=4.87 \times 10^{14} \mathrm{~s}^{-1}$

Frequency of emission $(v)=4.87 \times 10^{14} \mathrm{~s}^{-1}$

(b) Velocity of radiation, $(c)=3.0 \times 10^{8} \mathrm{~ms}^{-1}$

Distance travelled by this radiation in $30 \mathrm{~s}$

$=\left(3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)(30 \mathrm{~s})$

$=9.0 \times 10^{9} \mathrm{~m}$

(c) Energy of quantum $(E)=h v$

$\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(4.87 \times 10^{14} \mathrm{~s}^{-1}\right)$

Energy of quantum $\left.(E)=32.27 \times 10^{-20}\right]$

(d) Energy of one photon (quantum) $=32.27 \times 10^{-20} \mathrm{~J}$

Therefore, $32.27 \times 10^{-20} \mathrm{~J}$ of energy is present in 1 quantum.

Number of quanta in $2 \mathrm{~J}$ of energy

$=\frac{2 \mathrm{~J}}{32.27 \times 10^{-20} \mathrm{~J}}$

$=6.19 \times 10^{18}$

$=6.2 \times 10^{18}$