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.
 

1.	Cesium
2.	Potassium
3.	Sodium
4.	Lithium

The correct ranking of the targets according to their work function first will be:

1.	(i) > (ii) > (iii) > (iv)	2.	(i) > (iii) > (ii) > (iv)
3.	(iv) > (iii) > (ii) > (i)	4.	(i) = (iii) > (ii) = (iv)

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. \(2\)

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)

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. \(\lambda_0\)