Question

A sodium street light gives off yellow light that has a wavelength of $600$ nm. Then
(For energy of a photon take E = $\frac{12400eV\stackrel{\circ }{\text{A}}}{\lambda \left(\stackrel{\circ }{\text{A}}\right)}$

• A. Frequency of this light is $7\times {10}^{14}$ $s^{-1}$
• B. Frequency of this light is $5\times {10}^{14}$ $s^{-1}$
• C. Wavenumber of this light is $3\times {10}^6$ $m^{-1}$
• D. Energy of the photon is approximaltely 2.07 eV

Answer 1

Answer: (B), (D)

The correct options are
B Frequency of this light is $5\times {10}^{14}$ $s^{-1}$
D Energy of the photon is approximaltely 2.07 eV

Given: wavelength = 600 nm = $600\times {10}^{-9}$ m = 6000 $\stackrel{\circ }{\text{A}}$
We know, $\nu =\frac{c}{\lambda }$
where c = speed of light
$\nu$ = frequency of the light
$\lambda$ = wavelength of the radiation
So, $\nu$ = $\frac{3\times {10}^8}{600\times {10}^{-9}}$ = $5\times {10}^{14}$ $s^{-1}$
E = $\frac{12400eV\stackrel{\circ }{\text{A}}}{6000\stackrel{\circ }{\text{A}}}$

E = 2.07 $eV$