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JEE Main 2026 (5 Apr, Shift 1) · Transitions, Photon Emission & Excitation

Transitions, Photon Emission & Excitation question from JEE Main 2026 (5 Apr, Shift 1)

Q7JEE Main 2026 · 5 APR · SHIFT 1Medium

Transitions, Photon Emission & Excitation

In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit (f)(f). The ratio of the radius of the orbits in given by ri:rf=16:4r_{i} : r_{f} = 16 : 4. The wavelength of photon emitted due to this transition is ______ nm.

(Given Rydberg constant =1.0973×107 /m= 1.0973 \times 10^{7}\text{ /m})

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Answer: (c)

  1. r∝n2r \propto n^2, so ri:rf=16:4r_i : r_f = 16 : 4 gives ni=4n_i = 4, nf=2n_f = 2.
  2. 1λ=R(14−116)=3R16\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}
  3. λ=163×1.0973×107=4.86×10−7\lambda = \frac{16}{3 \times 1.0973 \times 10^{7}} = 4.86 \times 10^{-7} m
  4. So the answer is (c) 486 nm.
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