Question

When an excited electron falls from n = 3 to n = 2, a red radiation in the Balmer series emitted. What is the frequency of this photon?
A. 6.63 x 10-34 Hz
B. 2.19 x 1016 Hz
C. 5.28 x 1011 Hz
D. 4.57 x 1014 Hz

Answer 1

Answer:

The frequency of this photon is $4.57\times10^{14}\ Hz$

(D) is correct option.

Explanation:

Given that,

Excited states,

$n=3$

$n=2$

We need to calculate the wavelength

Using formula for energy

$E=13.6(\dfrac{1}{n^2}-\dfrac{1}{m^2})$

$E=13.6(\dfrac{1}{4}-\dfrac{1}{9})$

$E=1.888\ eV$

$E=1.888\times1.6\times10^{-19}\ J$

$E=3.0208\times10^{-19}\ J$

We need to calculate the frequency

Using formula of frequency

$E=hf$

$f=\dfrac{E}{h}$

Where, E =energy

$f=\dfrac{3.0208\times10^{-19}}{6.63\times10^{-34}}$

$f=4.57\times10^{14}\ Hz$

Hence, The frequency of this photon is $4.57\times10^{14}\ Hz$