Question

A sodium street light gives off yellow light that has a wavelength of 600 nm. Then
( h = $6.6\times {10}^{-34}$J.s)

• 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 the light is $3\times {10}^6{m}^{-1}$
• D. Energy of the photon is 5 eV

Answer 1

The correct option is B Frequency of this light is $5\times {10}^{14}{s}^{-1}$

Given,
Wavelength of the light = 600 nm
We know,
$h\nu =E$
$\therefore \text{Frequency}\nu =\frac{c}{\lambda }=\frac{3\times {10}^8}{600\times {10}^{-9}}=5\times {10}^{14}Hz$

Again,
$\text{wave number}\stackrel{-}{\nu }=\frac{1}{\lambda }=\frac{1}{600\times {10}^{-9}}=1.67\times {10}^6{m}^{-1}$

So energy of the photon,
$E=h\nu =(6.6\times {10}^{-34})(5\times {10}^{14})$ J
$=3.3\times {10}^{-19}J$
$=3.3\times {10}^{-19}\times 6.2\times {10}^{18}eV$ = 2.05 eV