1. Write the Bohr model energy relations

For the electron in the $n$th orbit of hydrogen:

$E_n=-\frac{13.6}{n^2}\,\text{eV}$

Using the Coulomb-force/virial relation for Bohr orbit:

$K_n=-E_n=\frac{13.6}{n^2}\,\text{eV}$

and

$P_n=2E_n=-\frac{27.2}{n^2}\,\text{eV}$

So:

• $K_n$ is positive and proportional to $\dfrac{1}{n^2}$
• $P_n$ is negative and proportional to $-\dfrac{1}{n^2}$
• $E_n$ is negative and proportional to $-\dfrac{1}{n^2}$

2. Consider transition to a higher level

A transition to a higher level means:

$n \to n' \quad \text{with} \quad n'>n$

Hence,

$\frac{1}{n'^2}<\frac{1}{n^2}$

Now check each quantity.

3. Change in kinetic energy $K$

Since

$K_n=\frac{13.6}{n^2}$

when $n$ increases, $K$ decreases.

So, kinetic energy decreases.

4. Change in potential energy $P$

Since

$P_n=-\frac{27.2}{n^2}$

for higher $n$, the value becomes less negative.

Example:

$P_1=-27.2\,\text{eV}, \qquad P_2=-6.8\,\text{eV}$

Clearly,

$-6.8 > -27.2$

So, potential energy increases.

5. Change in total energy $E$

Since

$E_n=-\frac{13.6}{n^2}$

for higher $n$, the total energy also becomes less negative.

Example:

$E_1=-13.6\,\text{eV}, \qquad E_2=-3.4\,\text{eV}$

Thus,

$-3.4 > -13.6$

So, total energy increases.

6. Match with options

• $K$ decreases
• $P$ increases
• $E$ increases

Therefore, the correct option is:

$\boxed{\text{B}}$