JEE Advanced 2015 · Paper 1 · Physics Q2

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PhysicsInteger+4Question 2 of 60

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 ( $hc=1242$ eV nm)

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Answer and worked solution

Official answer: 2

Photon energy. $E=\dfrac{1242}{90}=13.8$ eV.

Binding energy. Ionisation needs $13.8-10.4=3.4$ eV.

Level. $\dfrac{13.6}{n^2}=3.4$ , so $n=2$ .

Answer: 2

Topics:Bohr model and hydrogen spectrum Photoelectric effect

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JEE Advanced 2015 · Paper 1, Physics question 2