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The ionization energy of a hydrogen-like Bohr atom is 4 Rdybergs. Find the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state:

[1Rydberg=2.2×1018joule]

[h=6.6×1034Js,c=3×108m/s.]

Bohr radius of hydrogen atom =5×1011m]

 

a
400A     °
b
300A     °
c
500A     °
d
600A     °

detailed solution

Correct option is B

The energy in the ground stateE1=−4 RydbergE1=−4×2.2×1018JThe energy of the first excited state (n=2)E2=E14=-2.2×10−18JThe energy differenceΔE=E2−E1=3×2.2×10−18JNow, the wavelength of radiation emitted isλ=hcΔEλ=6.6×10−34×3×1083×2.2×10−18=300A    ∘

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