Neon gas is generally used in the sign boards.

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

radiation

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

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

$=6.19 \times 10^{18}$

$=6.2 \times 10^{18}$