The momentum of a photon of frequency $5 \times {10^{17}}{{\rm{s}}^{ - 1}}$ is nearly:

$1.1 \times {10^{ - 24}}{\rm{\;kg\;m\;}}{{\rm{s}}^{ - 1}}$

$3.33 \times {10^{ - 43}}{\rm{\;kg\;m\;}}{{\rm{s}}^{ - 1}}$

$2.27 \times {10^{ - 40}}{\rm{\;kg\;m\;}}{{\rm{s}}^{ - 1}}$

$2.27 \times {10^{ - 38}}{\rm{\;kg\;m\;}}{{\rm{s}}^{ - 1}}$

JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

1 Hard

1 of 1

Application +4 | -1

The momentum of a photon of frequency

5 \times {10^{17}}{\mathrm{s}^{ - 1}}

is nearly:

View answer and explanation

Correct option: A

1.1 \times {10^{ - 24}}\mathrm{\;kg\;m\;}{\mathrm{s}^{ - 1}}

Option A (

1.1 \times {10^{ - 24}}\mathrm{\;kg\;m\;}{\mathrm{s}^{ - 1}}

) is correct because the momentum of a photon is calculated using the formula

p = h/\lambda = hv/c

Practice All Chapter MCQs →

n^{\text {th }}

\mathrm{A}

\times 10^{12}

\frac{1}{{ }^{\prime} \mathrm{B}^{\prime}}

\ell = 1,\,\,m = 0

1.25 \times {10^{ - 13}} \times {A^{1/3}}\mathrm{cm}.