A.

frequency

h - Planck's constant

E - Energy of photon

$$\begin{gathered} \mathrm{E} = \mathrm{h\nu } \\ \Rightarrow \left[\mathrm{h}\right] = \frac{\left[\mathrm{E}\right]}{\left[\mathrm{\nu }\right]} = \frac{\left[\mathrm{M} {\mathrm{L}}^2 \mathrm{T}{ }^{-2}\right]}{\left[{\mathrm{T}}^{ -1}\right]} \\ = \left[\mathrm{M} {\mathrm{L}}^2 {\mathrm{T}}^{ -1}\right] \end{gathered}$$

And I = moment of inertia = MR²

$$\begin{gathered} \Rightarrow \left[\mathrm{I}\right] = \left[\mathrm{M}\right] \left[{\mathrm{L}}^2\right] = \left[{\mathrm{ML}}^2\right] \\ \mathrm{Hence} , \\ \frac{\left[\mathrm{h}\right]}{\left[\mathrm{I}\right]} = \frac{\left[\mathrm{M} {\mathrm{L}}^2{\mathrm{T}}^{ -1}\right]}{\left[{\mathrm{T}}^{ -1}\right]} \\ = \left[{\mathrm{T}}^{ -1}\right] \\ = \frac{1}{\left[\mathrm{T}\right]}\Rightarrow \mathrm{dimension} \mathrm{of} \mathrm{frequency} \end{gathered}$$

Alternative method:-

$$\begin{gathered} \frac{\mathrm{h}}{\mathrm{I}} =\frac{{\displaystyle \raisebox{1ex}{$\mathrm{E}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\nu }$}\right.}}{\mathrm{I}} \\ = \frac{\mathrm{E} \times \mathrm{T}}{\mathrm{I}} =\frac{\left(\mathrm{kg} {\mathrm{m}}^2/{\mathrm{s}}^2\right){\displaystyle }{\displaystyle \times }{\displaystyle }{\displaystyle \mathrm{s}}}{\left(\mathrm{kg} {\mathrm{m}}^2\right)} \\ = \frac{1}{\mathrm{s}} = \frac{1}{\mathrm{time}} = \mathrm{freauency} \end{gathered}$$

Thus dimension of h/I is same as frequency.

D.

8 × 10⁶ m/s

The solution to our problem consists in Einstein's photoelectric equation. 

Einstein's photoelectric equation can be written as

$$\begin{gathered} \frac{1}{2} {\mathrm{mv}}^2 = \mathrm{h\nu } - \mathrm{\varphi } \\ \frac{1}{2}\mathrm{m}\times {\left(4 \times {10}^6\right)}^2 = 2{\mathrm{h\nu }}_{\mathrm{o}} - {\mathrm{h\nu }}_{\mathrm{o}} ...\left(\mathrm{i}\right) \\ \mathrm{and} \\ \frac{1}{2}\times \mathrm{m} \times {\mathrm{v}}^2 = 5 {\mathrm{h\nu }}_{\mathrm{o}}- {\mathrm{h\nu }}_{\mathrm{o}} ......\left(\mathrm{ii}\right) \end{gathered}$$

Dividing equation (ii) by eq (i)

$$\begin{gathered} \Rightarrow \frac{{\mathrm{\nu }}^2}{{\left(4 \times {10}^6\right)}^2} = \frac{4{\mathrm{h\nu }}_{\mathrm{o}}}{{\mathrm{h\nu }}_{\mathrm{o}}} \\ \Rightarrow {\mathrm{\nu }}^2 = 4 \times 16 \times {10}^{12} \\ \therefore {\mathrm{v}}^2 = 64 \times {10}^{12} \\ \mathrm{\nu } = 8 \times {10}^6 \mathrm{m}/\mathrm{s} \end{gathered}$$

The efficiency of photoelectric effect is less than 1 o .ie, number of photons less than 1 % are capable of ejecting photoelectrons.

A.

< 124 kV

From conservation of energy the kinetic energy of electron equals the maximum photon energy (we neglect the he work function $\mathrm{\varphi }$ because it is normally so small compared to eV_o ). 

       eV_o = h v_max

         $$\begin{gathered} {\mathrm{eV}}_{\mathrm{o}} = \frac{\mathrm{hc}}{{\mathrm{\lambda }}_{\min }} \\ \therefore {\mathrm{V}}_{\mathrm{o}} = \frac{\mathrm{hc}}{{\mathrm{e\lambda }}_{\min }} \\ \Rightarrow {\mathrm{V}}_{\mathrm{o}} = \frac{12400 \times {10}^{-10}}{{10}^{-11}} \end{gathered}$$

                 = 124 kV

Hence, accelerating voltage for electrons in X-ray machine should be less than 124 kV.

B.

frequency

In photoelectric effect for a given photosensitive material, there exists a certain minimum cut-off frequency, called the threshold frequency, below which no emission of photoelectrons takes place no matter how intense the light is.

B.

black

If an object reflects the colour of light incident on it, it will appear with that colour but if object absorbs the colour of light, it will appear to be black. since the given wavelength does not belong to green, so it will be absorbed by the leaf and hence, it would appear to be black.