A single electron species in energy level (orbit number n ) with energy X was provided with excess of energy so that it jumps to higher energy level with energy Y. If it can emit radiations of six dif

Question

A single electron species in energy level (orbit number n ) with energy X was provided with excess of energy so that it jumps to higher energy level with energy Y. If it can emit radiations of six different wavelengths on de-excitement between these two energy levels, then the correct relation is

TestHub · Accepted Answer

The energy of an electron in a single-electron species is $E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}}$ eV.

Let the initial energy level be $n_{1} = n$ and its energy be $X$ . So, $X = - 13.6 \frac{Z ^{2}}{n ^{2}}$ .

Let the final energy level be $n_{2}$ and its energy be $Y$ . So, $Y = - 13.6 \frac{Z ^{2}}{n _{2}^{2}}$ .

The number of possible radiations emitted when an electron de-excites from $n_{2}$ to $n_{1}$ is given by the formula:

Number of lines $= \frac{( n _{2} - n _{1} ) ( n _{2} - n _{1} + 1 )}{2}$

Given that the number of radiations is 6, and $n_{1} = n$ :

$6 = \frac{( n _{2} - n ) ( n _{2} - n + 1 )}{2}$

$12 = (n_{2} - n) (n_{2} - n + 1)$

Let $k = n_{2} - n$ .

$12 = k (k + 1)$

By inspection, $k = 3$ satisfies the equation ( $3 \times 4 = 12$ ).

So, $n_{2} - n = 3$ , which means $n_{2} = n + 3$ .

Now, let's find the ratio $\frac{X}{Y}$ :

$\frac{X}{Y} = \frac{- 13.6 \frac{Z ^{2}}{n ^{2}}}{- 13.6 \frac{Z ^{2}}{n _{2}^{2}}} = \frac{n _{2}^{2}}{n ^{2}} = (\frac{n _{2}}{n})^{2}$

Substitute $n_{2} = n + 3$ :

$\frac{X}{Y} = (\frac{n + 3}{n})^{2} = (1 + \frac{3}{n})^{2}$

Taking the square root of both sides:

$\frac{X}{Y} = 1 + \frac{3}{n}$

This matches option C.

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