Question

Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from n=7 to n=1

Answer 1

Answer : The energy of photon = $21.369\times 10^{-19}J$

Solution : Given,

Final energy level = 1

Initial energy level = 7

Formula used :

$\frac{1}{\lambda}=R[ \frac{1}{n^2_{final}}- \frac{1}{n^2_{initial}}]$    ..........(1)

where,

$\lambda$ = wavelength

R = Rydberg's constant = 1.0974 $\times 10^7m^{-1}$

$n_{final}$ = final energy level

$n_{initial}$ = initial energy level

Now put all the given values in above formula (1), we get the value of wavelength.

$\frac{1}{\lambda}=1.0974\times 10^7m^{-1} [\frac{1}{1^2}- \frac{1}{7^2}]$ =  $1.0750\times 10^7m^{-1}$

$\lambda=0.9302\times 10^{-7}m$

Now we have to calculate the energy of photon by using formula,

$E=\frac{h\times c}{\lambda}$

where,

E = energy

h = planck's constant = $6.626\times 10^{-7}Js$

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

$\lambda$ = wavelength

put all the values in this formula, we get the energy of photon.

$E=\frac{6.626\times 10^{-7}Js\times 3\times 10^8m/s}{0.9302\times 10^{-7}m}$ = $21.369\times 10^{-19}J$