JEE Main · PhysicsHard

A hydrogen-like atom with atomic number $Z$ is in an excited state of principal quantum number $n$. It emits a photon of wavelength $\lambda$ when it transitions to the ground state. If the recoil momentum of the atom is $p$, then the value of $Z$ is given by which expression?

  1. A.$Z = \frac{p \lambda}{h} \left( \frac{n^2}{n^2 - 1} \right)$
  2. B.$Z = \frac{h}{p \lambda} \left( \frac{n^2 - 1}{n^2} \right)$
  3. C.$Z = \frac{p \lambda}{h} \left( \frac{n^2 - 1}{n^2} \right)$
  4. D.$Z = \frac{h}{p \lambda} \left( \frac{n^2}{n^2 - 1} \right)$
Show correct answer & step-by-step solution

Correct answer: A$Z = \frac{p \lambda}{h} \left( \frac{n^2}{n^2 - 1} \right)$

Solution

The momentum of the emitted photon is . Using the Rydberg formula for energy transition, the energy of the photon is , and equating leads to the result.

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A hydrogen-like atom with atomic number Z is in an excited state of principal qu… — JEE Main Physics Question with Solution | Solv