Question

A light bulb consumes $15\text{W}$ of power. $10\text{W}$ is lost as heat energy. The bulb emits the number of photons per second is $1.26\times {10}^{19}{\text{s}}^{-1}$. The wavelength of the light is (nearly)
$[\text{Use}hc=1240\text{eV nm}]$

• A.
  $280\text{nm}$
• B.
  $600\text{nm}$
• C.
  $500\text{nm}$
• D.
  $780\text{nm}$

Answer 1

The correct option is C

$500\text{nm}$
Given,

$P=15\text{W};\text{Energy loss}=10\text{W}\frac{n}{t}=1.26\times {10}^{19}{\text{s}}^{-1}$

$\text{The light energy emitted/sec}=P-\text{Energy loss/sec}$

$=15\text{W}-10\text{W}=5\text{W}$

The energy of each photon is,

$E=\frac{P}{\frac{n}{t}}=\frac{5}{1.26\times {10}^{19}}=3.96\times {10}^{-19}\text{J}=\frac{3.96\times {10}^{-19}}{1.6\times {10}^{-19}}\text{eV}$

The wavelength of the light emitted is,

$\lambda =\frac{hc}{E}$

$=\frac{1240\text{eV (nm)}\times 1.6\times {10}^{-19}}{3.96\times {10}^{-19}\text{eV}}[\because hc=1240\text{eV (nm)}]$

$\therefore \lambda \approx 500\text{nm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.