Question

Given that the energy difference between the ground state and the first excited electronic state (E) for the sodium atom is 3.373 × 10-19 J, calculate the frequency, , corresponding to a photon possessing this energy. Next, calculate the wavelength (in nm) for this photon.

Answer 1

Answer: Frequency corresponding to a photon possessing this energy is $0.5\times 10^{15}Hertz$

Wavelength for this photon is 600 nm.

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

$E=\frac{hc}{\lambda}$

E= energy

$\lambda$ = wavelength of the wave

h = Planck's constant  = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$3.373\times 10^{-19}=\frac{6.626\times 10^{-34}\times 3\times 10^8m/s}{\lambda}$

$\lambda=6\times 10^{-7}m=600nm$   $1m=10^9nm$

Thus wavelength for this photon is 600 nm.

The relationship between wavelength and frequency of the wave follows the equation:

$\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency of the wave

c = speed of light  

$\nu=\frac{3\times 10^8m/s}{6\times 10^{-7}m}$

$\nu=0.5\times 10^{15}s^{-1}=0.5\times 10^{15}Hertz$

Thus frequency corresponding to a photon possessing this energy is $0.5\times 10^{15}Hertz$