Question

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

Answer 1

Answer : The energy of one photon of hydrogen atom is,
`3.03* 10^(-19)J`

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

`(1)/(\lambda)=R_H\left((1)/(n_i^2)-(1)/(n_f^2) \right )`

Where,

`\lambda` = Wavelength of radiation

`R_H` = Rydberg's Constant = 10973731.6 m⁻¹

`n_f` = Higher energy level = 3

`n_i`= Lower energy level = 2

Putting the values, in above equation, we get:

`(1)/(\lambda)=(10973731.6)\left((1)/(2^2)-(1)/(3^2) \right )`

`\lambda=6.56* 10^(-7)m`

Now we have to calculate the energy.

`E=(hc)/(\lambda)`

where,

h = Planck's constant =
`6.626* 10^(-34)Js`

c = speed of light =
`3* 10^8m/s`

`\lambda` = wavelength =
`6.56* 10^(-7)m`

Putting the values, in this formula, we get:

`E=((6.626* 10^(-34)Js)* (3* 10^8m/s))/(6.56* 10^(-7)m)`

`E=3.03* 10^(-19)J`

Therefore, the energy of one photon of hydrogen atom is,
`3.03* 10^(-19)J`