Correct answer is option A

Let speed of photon electron in first case is 2v

then in the second case speed is v.

For first case

$\frac{hc}{{\lambda }_1}=\varphi +\frac{1}{2}$m(2v)²

For second case,

$\frac{hc}{{\lambda }_2}=\varphi +\frac{1}{2}$mv²

$\therefore$   $\frac{\frac{hc}{{\lambda }_1}-\varphi }{\frac{hc}{{\lambda }_2}-\varphi }=\frac{\frac{1}{2}m\times 4v^2}{\frac{1}{2}m{v}^2}$

$\Rightarrow$   $\frac{hc}{{\lambda }_1}-\varphi$ = 4$\left(\frac{hc}{{\lambda }_2}-\varphi \right)$

$\Rightarrow$   $\frac{4hc}{{\lambda }_2}$ $-$ $\frac{hc}{{\lambda }_1}$ = 3$\varphi$

$\Rightarrow$   $\varphi$ $=$ $\frac{hc}{3}\left(\frac{4}{{\lambda }_2}-\frac{1}{{\lambda }_1}\right)$

$\Rightarrow$   $\varphi$ $=$ $\frac{1240}{3}\left(\frac{4}{540}-\frac{1}{350}\right)$

$\Rightarrow$   $\varphi$ = 1.8 eV

Correct answer is option C

The given maximum velocity is v and frequency is n. We know that the kinetic energy is given by

$KE=\frac{1}{2}m{v}^2=hn-\varphi$

where h is Planck's constant and $\varphi$ is the work function. Therefore, the kinetic energy of the incident light is

$E_1=hn-\varphi$ ..... (1)

When the frequency of the incident light is increased to 3n, then the kinetic energy is given by

$\frac{1}{2}m{v}_1^2=3hn-\varphi$

$\Rightarrow E_2=3hn-\varphi$ ..... (2)

Substituting $hn=E_1+\varphi$ [from Eq. (1)] in Eq. (2), we get

$E_2=3(E_1+\varphi )-\varphi$

$\Rightarrow E_2=3E_1+2\varphi$

$\Rightarrow \frac{1}{2}m{v}_1^2=3\times \frac{1}{2}m{v}^2+2\varphi$

$\Rightarrow v_1^2=3v^2+2\varphi \times \frac{2}{m}$

$\Rightarrow v_1^2=3v^2+\frac{4\varphi }{m}$

Thus, the velocity is more than $\sqrt{3}v$.

Correct answer is option B

We know,

Einstein's photo electric equation,

(KE)_max = $\frac{hc}{\lambda }$ $-$ $\varphi$₀

In first case,

K = $\frac{hc}{\lambda }$ $-$ $\varphi$₀           . . .(1)

In second case,

3K = $\frac{2hc}{\lambda }$ $-$ $\varphi$₀           . . .(2)

$\Rightarrow$   $3\left(\frac{hc}{\lambda }-{\varphi }_0\right)$ = $\frac{2hc}{\lambda }-{\varphi }_0$

$\Rightarrow$   $\frac{3hc}{\lambda }$ $-$ 3$\varphi$₀ = $\frac{2hc}{\lambda }$ $-$ $\varphi$₀

$\Rightarrow$   $\frac{hc}{\lambda }$ = 2$\varphi$₀

$\Rightarrow$   $\varphi$  =  $\frac{hc}{2\lambda }$

Correct answer is option C

We know,

Einstein's photoelectric equation,

$eV=\frac{hc}{\lambda }-{\varphi }_0$

and ${\varphi }_0$, $\frac{hc}{{\lambda }_0}$, where ${\lambda }_0$ is the threashhold wavelength.

$\therefore$   In first case,

eV $=\frac{hc}{{\lambda }_1}-\frac{hc}{{\lambda }_0}$      . . .(1)

and in second case,

3 eV $=\frac{hc}{{\lambda }_2}-\frac{hc}{{\lambda }_0}$       . . .(2)

Now, let slopping potential = V₁ when light of wavelength $\lambda$₃ is used then,

eV₁ $=\frac{hc}{{\lambda }_3}-\frac{hc}{{\lambda }_0}$       . . .(3)

From (1) and (2) get,

$3\left(\frac{hc}{{\lambda }_1}-\frac{hc}{{\lambda }_0}\right)=\frac{hc}{{\lambda }_2}-\frac{hc}{{\lambda }_0}$

$\Rightarrow$   $\frac{3hc}{{\lambda }_1}$ $-$ $\frac{hc}{{\lambda }_2}$ $=$ $\frac{2hc}{{\lambda }_0}$

$\Rightarrow$   $\frac{hc}{{\lambda }_0}$ $=$ $\frac{3hc}{2{\lambda }_1}$ $-$ $\frac{hc}{2{\lambda }_2}$

Putting this value of $\frac{hc}{{\lambda }_0}$ in equation 3,

eV₁ $=\frac{hc}{{\lambda }_3}-\frac{3hc}{2{\lambda }_1}+\frac{hc}{2{\lambda }_2}$

$\Rightarrow$   V₁ $=\frac{hc}{e}$ $\left[\frac{1}{{\lambda }_3}-\frac{3}{2{\lambda }_1}+\frac{1}{2{\lambda }_2}\right]$

Correct answer is option D

As $\lambda$ is increased, there will be a value of $\lambda$ above which photo electrons will be cases to come out so photo current will become zero. Hence $(d)$ is correct answer.