JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Modern Physics

Consider a hydrogen atom with its electron in the

n^{th}

orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then the value of

n

is (

h c = 1242 eV nm

)

Answer: 2

Solution

The energy of the incident photon is

E_{p h o t o n} = \frac{h c}{λ} = \frac{1242}{90} = 13.8

eV. The ionization energy from the

n^{t h}

level is

E_{n} = \frac{13.6}{n ^{2}}

eV. By energy conservation:

E_{p h o t o n} = E_{n} + K E_{e l ec t r o n}

, so

13.8 = \frac{13.6}{n ^{2}} + 10.4

. This gives

\frac{13.6}{n ^{2}} = 3.4

, hence

n^{2} = 4

and

n = 2

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Integer answer

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