JEE Main 2026 · 5 Apr · Shift 1 · Physics Q36

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PhysicsSingle correct+4 / -1Question 36 of 75

In the hydrogen atom, the electron makes a transition from the higher orbit ( $i$ ) to a lower orbit ( $f$ ). The ratio of the radius of the orbits in given by $r_i:r_f=16:4$ . The wavelength of photon emitted due to this transition is ________ nm.
(Given Rydberg constant $=1.0973\times10^7$ /m)

121

242

486

974

Answer and worked solution

Official answer: C

Orbits. $r\propto n^2$ , and $16:4$ gives $n_i=4$ , $n_f=2$ .

Wavelength.
$\frac1\lambda=R\left(\frac14-\frac1{16}\right)=\frac{3R}{16},\qquad\lambda=\frac{16}{3(1.0973\times10^7)}\text{ m}\approx486\text{ nm}.$

(C)

Topics:Bohr model and hydrogen spectrum

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JEE Main 2026 · 5 Apr · Shift 1, Physics question 36