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Q.

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 =2.2×10−18joule][h=6.6×10−34Js,c=3×108m/s.]Bohr radius of hydrogen atom =5×10−11m]

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a

400A     °

b

300A     °

c

500A     °

d

600A     °

answer is B.

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Detailed Solution

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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