Question

An electron of $\mathrm{H}$-atom de-exits from energy level $n_1=2$ to ${\mathrm{n}}_2=1$ and the emitted photon is incident on ${\mathrm{He}}^{+}$ ion in ground state and first excited states which of following transition is possible.

• A. $\mathrm{n}=1$ to $\mathrm{n}=4$
• B. $\mathrm{n}=2$ to $\mathrm{n}=4$
• C. $\mathrm{n}=2$ to$\mathrm{n}=3$
• D. $\mathrm{n}=3$ to $\mathrm{n}=4$

Answer 1

The correct option is B

$\mathrm{n}=2$ to $\mathrm{n}=4$

The above question can be solved as:

Step 1:

According to the Rydberg equation for Hydrogen like the atom:

Now, ${\mathrm{H}}^{+}$ according to the equation:

$\mathrm{E}=-13.6\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)\mathrm{ev}$

$=-13.6\times 1^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)ev$

$=-10.2ev$

Step 2:

Now ${\mathrm{He}}^{+}$ according to the question,

$-10.2=-13.6\times 2^2\left(\frac{1}{2^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$

$\frac{0.75}{4}=\left(\frac{1}{4}-\frac{1}{n_2^2}\right)$

$0.1875-0.25=-\frac{1}{n_2^2}$

or, $0.0625=\frac{1}{n_2^2}$

or, ${\mathrm{n}}_2^2=\frac{1}{0.0625}$

or, $n_2^2=16$

hence, $n_2=4$

So option (B) is correct.