Question

For the electronic transition from n = 2 to n = 4 in the hydrogen atom. a) Calculate the energy. b) Calculate the wavelength (in nm).

Answer 1

Answer :

(a) The energy is, $4.088\times 10^{-19}J$

(b) The wavelength (in nm) is, $4.862\times 10^2nm$

Explanation :

The general formula for the wavelength of spectral line emitted by a hydrogen atom, when it makes a transition from $n_2$ shell to $n_1$ shell is,

$\frac{1}{\lambda}=R_H[\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2}]$

where,

$\lambda$ = wavelength

Rydberg's Constant = $R_H=1.097\times 10^7m^{-1}$

In this, $n_1=2$ and $n_2=4$

$\frac{1}{\lambda}=(1.097\times 10^7)\times [\frac{1}{(2)^2}-\frac{1}{(4)^2}]$

$\lambda=4.862\times 10^{-7}m=4.862\times 10^2nm$

conversion used : $(1m=10^9nm)$

Now we have to calculate the energy.

Formula used :

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

$\lambda$ = Wavelength = $4.862\times 10^{-7}m$

E = energy = ?

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

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

Now put all the given values in the above formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{4.862\times 10^{-7}m}$

$E=4.088\times 10^{-19}J$

Therefore, the energy is $4.088\times 10^{-19}J$