For photoelectric emission from certain metals, the cutoff frequency is \(\nu\). If radiation of frequency \(2\nu\)$\mathrm{}$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be:
(\(m\) is the electron mass)

1.	\(\sqrt{\dfrac{h\nu}{m}}\)	2.	\(\sqrt{\dfrac{2h\nu}{m}}\)
3.	\(2\sqrt{\dfrac{h\nu}{m}}\)	4.	\(\sqrt{\dfrac{h\nu}{2m}}\)

A \(5\) W emits monochromatic light of wavelength \(5000~\mathring{A}\). When placed \(0.5\) m away, it liberates photoelectrons from a photosensitive metallic surface. When the source is moved \(1.0\) m away, the number of photoelectrons liberated is reduced by a factor of?
1. \(4\)
2. \(8\)
3. \(16\)
4. \(2\)

When the light of frequency \(2\nu_0\) (where \(\nu_0\) is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is \(v_1\). When the frequency of the incident radiation is increased to \(5\nu_0,\) the maximum velocity of electrons emitted from the same plate is \(v_2.\) What will be the ratio of \(v_1\) to \(v_2\)?

An electron of mass m with an initial velocity \(\overrightarrow v= v_0\hat i\)$\stackrel{}{}$\( ( v_o > 0 ) \) enters in an electric field \(\overrightarrow E = -E_0 \hat i\)\((E_0 = \text{constant}>0)\) at \(t=0\). If \(\lambda_0\)${\mathrm{}}_{}$, is its de-Broglie wavelength initially, then what will be its de-Broglie wavelength at time \(t\)?
1. \(\frac{\lambda_0}{\left(1+ \frac{eE_0}{mv_0}t\right)}\)
2. \(\lambda_0\left(1+ \frac{eE_0}{mv_0}t\right)\)
3. \(\lambda_0 t\)
4. \(\lambda_0\)

The figure shows different graphs between stopping potential \(V_0\) and frequency (\(\nu\)) for the photosensitive surfaces of cesium, potassium, sodium and lithium. The plots are parallel.
 

The value of stopping potential in the following diagram is given by:
    

The stopping potential for photoelectrons: