JEE Main 2022 · 25 Jul · Shift 2 · Physics Q50

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PhysicsSingle correct+4 / -1Question 50 of 90

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength $\lambda$ . The value of principal quantum number 'n' of the excited state will be : (R : Rydberg constant)

\sqrt{\dfrac{\lambda R}{\lambda-1}}

\sqrt{\dfrac{\lambda R}{\lambda R-1}}

\sqrt{\dfrac\lambda{\lambda R-1}}

\sqrt{\dfrac{\lambda R^2}{\lambda R-1}}

Answer and worked solution

Official answer: B

Rydberg. $\frac1\lambda=R\left(1-\frac1{n^2}\right)$ , so $n=\sqrt{\dfrac{\lambda R}{\lambda R-1}}$ .

(B)

Topics:Bohr model and hydrogen spectrum

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JEE Main 2022 · 25 Jul · Shift 2, Physics question 50