Correct option is (A)

$\begin{array}{rl}& \text{ K.E. }=\mathrm{hf}-\varphi \\ & \tan \theta =\mathrm{h}\end{array}$

Correct option is (C)

$\begin{array}{rl}& \text{ For P.E.E. : }\lambda \le \frac{hc}{W_e}\\ & \lambda \le \frac{1240\mathrm{nm}-\mathrm{eV}}{3\mathrm{eV}}\\ & \lambda \le 413.33\mathrm{nm}\\ & {\lambda }_{max}\approx 414\mathrm{nm}\text{ for P.E.E. }\end{array}$

Correct option is (C)

$1.6\times {10}^{15}\mathrm{Hz}$

Correct answer is 122

The Franck-Hertz experiment provides evidence for quantized energy levels within atoms. When atoms are excited by electrons with a specific kinetic energy, they can jump to higher energy levels. Upon returning to lower levels, they emit photons whose energies correspond to the difference between these levels. The first dip in the current-voltage graph for hydrogen, observed at $10.2\mathrm{V}$, corresponds to the energy required to excite a hydrogen atom to its first excitation level. The wavelength of the light emitted when the atom returns to its ground state can be calculated using the energy of the photon emitted.

To find the wavelength ($\lambda$) of light emitted, we use the relationship between energy ($E$), Planck's constant ($h$), the speed of light ($c$), and wavelength ($\lambda$), given in the equation form as $E=\frac{hc}{\lambda }$.

However, we are given $hc$ in electron volts per nanometer ($1245\mathrm{eV}\cdot \mathrm{nm}$), and the energy is also given in terms of voltage ($10.2\mathrm{V}$). First, we convert the energy into electron volts (eV) using the formula: $E=eV$, where $e$ is the charge of an electron ($1.6\times {10}^{-19}\mathrm{C}$).

The energy in electron volts can be directly calculated as:

$E=10.2\mathrm{V}\cdot 1.6\times {10}^{-19}\mathrm{C}/\mathrm{electron}=10.2\mathrm{eV}$,

since $1\mathrm{V}\cdot 1\mathrm{C}=1\mathrm{eV}$ by definition.

Next, using the energy-wavelength relationship and the given value for $hc$, the wavelength can be calculated as:

$\lambda =\frac{hc}{E}$

Substituting the given values yields:

$\lambda =\frac{1245\mathrm{eV}\cdot \mathrm{nm}}{10.2\mathrm{eV}}$

This simplifies to:

$\lambda =\frac{1245}{10.2}\mathrm{nm}$

Calculating this gives:

$\lambda \approx 122.06\mathrm{nm}$.

Therefore, the wavelength of light emitted by the hydrogen atom when excited to the first excitation level is approximately $122.06\mathrm{nm}$.