Q11 / 51 · JEE Advanced 2023 (Paper 2) · Chemistry

For He+, a transition takes place from the orbit of radius 105.8pm to the orbit of radius 26.45pm. The wavelength (in nm ) of the emitted photon during the transition is

Bohr radius, a=52.9 pm
Rydberg constant, RH=2.2×10-18 J
Planck's constant, h=6.6×10-34Js
Speed of light, c=3×108 ms-1 ]

Correct answer: 30

Solution

The radius of the nth orbit can be represented as,

r=52.9×n2zpm

∴  105.8=52.9×n22  ∴n2=2

and 26.45=52.9×n22  ∴n1=1

∵ΔE=RHhC×z21n12-1n22

hcλ=RHhC×z21n12-1n22

6.6×10-34×3×108λ=2.2×10-18×4×11-14

6.6×10-34×3×108λ=2.2×10-18×4×34

∴λ=300 A

∴λ=30 nm

Hence, the correct answer is 30.

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