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Question: Answered & Verified by Expert
An electron of an atom transits from $n_{1}$ to $n_{2}$. In which of the following maximum frequency of photon will be emitted?
PhysicsAtomic PhysicsJEE Main
Options:
  • A $n_{1}=1$ to $n_{2}=2$
  • B $n_{1}=2$ to $n_{2}=1$
  • C $n_{1}=2$ to $n_{2}=6$
  • D $n_{1}=6$ to $n_{2}=2$
Solution:
2772 Upvotes Verified Answer
The correct answer is: $n_{1}=2$ to $n_{2}=1$
For emission of energy, electron of H-atom must fall from higher energy state to lower energy state. Hence, options (a) and (c) are not possible.
In option (b), $n_{1}=2, n_{2}=1$
$\therefore$ Energy of emitted photon is given as
$$
\begin{aligned}
E^{\prime} &=-13.6 Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\
\Rightarrow \quad h v^{\prime} &=-13.6 Z^{2}\left(\frac{1}{2^{2}}-\frac{1}{1^{2}}\right) \\
v^{\prime} &=\frac{13.6 Z^{2}}{h} \times \frac{3}{4}
\end{aligned}
$$
In option (c), $n_{1}=2, n_{2}=6$, hence energy of emitted photon is given as
$$
\begin{aligned}
& E^{\prime \prime}=-13.6 Z^{2}\left(\frac{1}{6^{2}}-\frac{1}{2^{2}}\right) \\
\Rightarrow & h v^{\prime \prime}=13.6 Z^{2} \times \frac{8}{36} \\
\Rightarrow & \quad v^{\prime \prime}=\frac{13.6 Z^{2}}{h} \times \frac{8}{36}
\end{aligned}
$$
Hence, from Eqs. (i) and (ii), we get
$$
v^{\prime}>v^{\prime \prime}
$$

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