Question

A certain laser transition emits $6.37\times {10}^{15}{\text{quanta sec}}^{-1}{\text{m}}^{-2}$. Calculate the power output in ${\text{J sec}}^{-1}{\text{m}}^{-2}$.
Given: $\lambda =632.8\text{nm}$

• A. $2.5\times {10}^{-3}{\text{J sec}}^{-1}{\text{m}}^{-2}$
• B. $2.0\times {10}^{-3}{\text{J sec}}^{-1}{\text{m}}^{-2}$
• C. $1.6\times {10}^{-3}{\text{J sec}}^{-1}{\text{m}}^{-2}$
• D. $2.3\times {10}^{-3}{\text{J sec}}^{-1}{\text{m}}^{-2}$

Answer 1

The correct option is B $2.0\times {10}^{-3}{\text{J sec}}^{-1}{\text{m}}^{-2}$

We know,
$E=nhv=n\times \frac{hc}{\lambda }$
$\text{Power output}=\text{No. of quanta}\times \frac{hc}{\lambda }$
$=\frac{6.37\times {10}^{15}\times 6.62\times {10}^{-34}\times 3\times {10}^8}{632.8\times {10}^{-9}}$
$=1.999\times {10}^{-3}{\text{J sec}}^{-1}{\text{m}}^{-2}$
$=2\times {10}^{-3}{\text{J sec}}^{-1}{\text{m}}^{-2}$