Question

In a hydrogen atom, the electron is in nth excited state. It may come down to second excited state by emitting ten different wavelengths. What is the value of n:

• A. 6
• B. 7
• C. 8
• D. 5

Answer 1

Answer: (A)

6

Answer 2

The electron is in the nth excited state, so its principal quantum number is $N_i = n+1$. It de-excites to the second excited state, which corresponds to the principal quantum number $N_f = 2+1=3$. The number of distinct wavelengths emitted during transitions between $N_i$ and $N_f$ is given by $\frac{(N_i - N_f)(N_i - N_f + 1)}{2}$.

Substituting the values:

$10 = \frac{((n+1) - 3)((n+1) - 3 + 1)}{2}$

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

$20 = (n-1)(n-2)$

Let $x = n-2$. Then $x(x+1) = 20$.

$x^2 + x - 20 = 0$

$(x+5)(x-4) = 0$

Since $x = n-2$ must be positive (as $n-1$ and $n-2$ are consecutive positive integers), $x=4$.

$n-2 = 4 \implies n=6$.