An electron in an atom jumps in such a way that its kinetic energy changes from $x$ to $\frac{x}{4}$. The change in potential energy will be

What is the maximum number of emission lines when the excited electron of an $H$ atom in $n=6$ drops to the ground state?

Given below are two statements:
Statement $I$: Bohr's theory accounts for the stability and line spectrum of $Li^{+}$ ion.
Statement $II$: Bohr's theory was unable to explain the splitting of spectral lines in the presence of a magnetic field.
In the light of the above statements,choose the most appropriate answer from the options given below:

$A$ bulb of $40 \ W$ is producing light of wavelength $620 \ nm$ with $80 \%$ efficiency. The number of photons emitted by the bulb in $20 \ s$ is $(1 \ eV = 1.6 \times 10^{-19} \ J, hc = 12400 \ eV \ \mathring{A})$

When the electron passes from the energy state nearest to the nucleus to the third energy level,it will:

According to Bohr's theory,
$E_{n} = \text{Total energy}, K_{n} = \text{Kinetic energy}, V_{n} = \text{Potential energy}, r_{n} = \text{Radius of } n^{\text{th}} \text{ orbit}$
Match the following:

Column $I$	Column $II$
$A$. $V_{n} / K_{n} = ?$	$P$. $0$
$B$. If radius of $n^{\text{th}}$ orbit $\propto E_{n}^{x}, x = ?$	$Q$. $-1$
$C$. Angular momentum in lowest orbital	$R$. $-2$
$D$. $1/r_{n} \propto Z^{y}, y = ?$	$S$. $1$