Question

If an atom exists in the excited state n=5, the maximum number of transition takes place is :

Answer 1

Answer:

10

Explanation:

Answer 2

Answer:

If an atom exists in the excited state n = 5, the maximum number of transition that takes place is 10.

Explanation:

Given:

If an atom exists in the excited state n = 5.

To find:

the maximum number of transition that takes place.

From the $\mathrm{n}^{\text {th }}$ state, the electron may go to $(\mathrm{n}-1)^{\text {th }}$state, . . ., and state or  1st state. So there exists ( $\mathrm{n} 1$ ) possible transitions beginning from the $\mathrm{n}^{\mathrm{th}}$ state. The atoms reaching $(\mathrm{n}-1)^{\text {th }}$ state may create $\left(\mathrm{n}_{2}\right)$ various transitions. The atoms reaching $(\mathrm{n}-2)^{\text {th }}$ state may create (n 3) various transitions. Similarly for other lower states. During each transition, a photon with energy hv and wavelength $\lambda$ exists emitted out. Hence, the entire number of possible transitions exists equivalent to the number of photons emitted. The entire number of possible transitions exists

$(\mathrm{n}-1),(\mathrm{n}-2),(\mathrm{n}-3), \ldots \ldots \ldots \ldots, 3,2,1=\frac{\mathrm{n}(\mathrm{n}-1)}{2}$

Therefore, for the transition of an electron from a higher energy state $n=4$ to a lower energy state, $\mathrm{n}=1$ the number of photons emitted exists

$\frac{\mathrm{n}(\mathrm{n}-1)}{2}=\frac{5(5-1)}{2}=\frac{20}{2}=10$

If an atom exists in the excited state n = 5, the maximum number of transition that takes place exists at 10.

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