Question

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

• 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 $2.07eV$
• D. Both (b) and (c)

Answer 1

Answer: (D)

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

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

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

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

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

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

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

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

To calculate energy of a photon,

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