Question

A sodium street light gives off yellow light that has a wavelength of $600 nm$. Then:
$\left(\right.$ Energy of photon $\left.=\frac{12400 eV \mathring{A} }{\lambda( \mathring{A})}\right)$

• 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. Energy of photon is approximately $2.07 eV$
• D. Both (b) and (c)

Answer 1

Answer: (D)

Formula for energy of photon is already mentioned, i.e.

$=\frac{12400 eV \mathring{A}}{\lambda( A )}$

$E =\frac{12400 eV \mathring{A}}{600 \times 10( \mathring{A} )} (1 nm =10 \mathring{A})$

$hv =\frac{12400}{6000} eV$

$hv =\frac{124}{60} \times 1.6 \times 10^{-19} J$

$\left(1 eV =1.6 \times 10^{-19} J \right)$

$6.626 \times 10^{-34} Js \times v=\frac{124}{60} \times 1.6 \times 10^{-19} J$

$v=5 \times 10^{14} s ^{-1}$

To calculate energy of a photon,

$E =\frac{12400 eV \mathring{A}}{6000 \mathring{A} }=2.07 eV$