Question

Calculate the frequency of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level. Recall

En = –2.18 x 10–18 J(1/n2)

a. 3.08 x 1015/s
b. 1.03 x 108/s
c. 2.06 x 1014/s
d. 1.35 x 10–51/s
e. 8.22 x 1014/s

Answer 1

Answer: The frequency of the light is $3.088\times 10^{15}s^{-1}$

Explanation:

The equation used to calculate the energy for a transition, we use the equation:

$E_n=-2.18\times 10^{-18}J(\frac{1}{n^2})$

where,

n = principal energy level

Calculating the energy difference:

$\Delta E=E_4-E_1$

$\Delta E=-2.18\times 10^{-18}\left (\frac{1}{4^2}-\frac{1}{1^1}\right)\\\\\Delta E=2.044\times 10^{-18}J$

To calculate the energy of the light for a given frequency, we use the equation given by Planck, which is:

$E=h\nu$

where,

E = energy of the light  = $2.044\times 10^{-18}J$

h = Planck's constant = $6.62\times 10^{-34}Js$

$\nu$ = frequency of the light = ?

Putting values in above equation, we get:

$2.044\times 10^{-18}J=6.62\times 10^{-34}Js\times \nu\\\\\nu=\frac{2.044\times 10^{-18}J}{6.62\times 10^{-34}Js}=3.088\times 10^{15}s^{-1}$

Hence, the frequency of the light is $3.088\times 10^{15}s^{-1}$