Question

Maximum number of photons emitted by a bulb capable of producing monochromatic light of wavelength $550\, nm$ is ..........., if $100\, V$ and $1\, A$ is supplied for one hour.

• A. $1 \times 10^{24}$
• B. $5 \times 10^{24}$
• C. $2 \times 10^{24}$
• D. $5 \times 10^{23}$
• E. $6 \times 10^{24}$

Answer 1

Answer: (A)

$E =\frac{h c}{\lambda} $

$E =\frac{\left(6.626 \times 10^{-34} J s \right)\left(3.0 \times 10^{8}\right)}{550 \times 10^{-9}} $

$E =0.036 \times 10^{-17} J$

Each proton carries $0.036 \times 10^{-17}$ J of energy.

$P= IV =1 A \times 100\,V$

$P=100 \,W$

A $100\, W$ source means that it outputs $100\,J$ of energy every one hour. So, in one second, $100 \times 3600\, J$ of energy is liberated.

Number of photons emitted $=$ total energy output in one second/energy carried per photon

$n=\frac{100 \times 3600}{0.036 \times 10^{-17}}=1 \times 10^{24}$