A hydrogen atom in its ground state is irradiated by light of wavelength 1026\ Å . The electron gets excited to higher energy state n . Given the ground state energy is -13.6\ eV and hc = 1242\ eV , then the value of

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 3

The energy of the incident photon is given by

E = \dfrac{hc}{\lambda}

Substituting the given values with

\lambda = 102.6

nm:

E = \dfrac{1242}{102.6} \approx 12.1

eV

The energy of the electron in the excited state

n

is:

E_n = E_1 + E

E_n = -13.6 + 12.1 = -1.5

eV

The energy of the

n

-th state of a hydrogen atom is given by:

E_n = \dfrac{-13.6}{n^2}

eV

\dfrac{-13.6}{n^2} = -1.5

n^2 = \dfrac{13.6}{1.5} \approx 9

n = 3

Answer: 3

A hydrogen atom in its ground state is irradiated by light of...

Options

Correct answer

Step-by-step solution

More from Atomic Physics