Question

The number of photons emitted per nanosecond by a deuterium lamp (400 nm) having a power of 1 microwatt (rounded off to the nearest integer) is \(\underline{\hspace{2cm}}\).
Given: \( h = 6.626 \times 10^{-34} \, \text{kg m}^2 \text{s}^{-1}, c = 3.0 \times 10^8 \, \text{m s}^{-1} \).

Hint:

The number of photons emitted by a light source can be calculated by dividing the power of the source by the energy of a single photon.

Answer 1

Correct Answer: 2000

The energy of one photon is given by:
\[ E = \frac{hc}{\lambda} \] where \( \lambda = 400 \, \text{nm} = 4.0 \times 10^{-7} \, \text{m} \).
Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34}) \times (3.0 \times 10^8)}{4.0 \times 10^{-7}} = 4.97 \times 10^{-19} \, \text{J}. \] The power emitted is 1 microwatt = \( 1 \times 10^{-6} \, \text{W} \). The number of photons per second is:
\[ \text{Number of photons per second} = \frac{P}{E} = \frac{1 \times 10^{-6}}{4.97 \times 10^{-19}} = 2.01 \times 10^{12}. \] Since 1 second = 1 billion nanoseconds, the number of photons emitted per nanosecond is:
\[ \frac{2.01 \times 10^{12}}{10^9} = 2000. \] Thus, the number of photons emitted per nanosecond is \( 2000 \).