Question

In an experiment on photoelectric effect, the emitter and the collector plates are placed at a separation of 10 cm and are connected through an ammeter without any cell. A magnetic field B exists parallel to the plates. The work function of the emitter is 2.39 eV and the light incident on it has wavelengths between 400 nm and 600 nm. Find the minimum value of B for which the current registered by the ammeter is zero. Neglect any effect of space charge.

• A. 3.65 × 10^–5 T
• B. 5.48 × 10^–6 T
• C. 2.85 × 10^–5 T
• D. 7.24 × 10^–6 T

Answer 1

The correct option is C

2.85 × 10^–5 T

As shown the downward magnetic field $\overrightarrow{B}$, will bend the electron towards the right along a circular path of radius $r=\frac{mV}{qB}$ , while its speed v = $|\overrightarrow{V}|$ stays constant.

Given that the separation between the plates is d = 100 cms, for any radius r ≤ d, the electron will miss the collector plate, hence not contributing to the current. But in a photocurrent, not all electrons will be travelling at the same speed. The maximum velocity of a photo-electron is given by the photo-electric equation as –

$\frac{1}{2}m{V}_{max}^2$ = hv - w(m = mass of electron; w = work function)

Since the fastest electron will also have the largest radius of trajectory under a magnetic field $(\therefore r\alpha v)$, its safe to claim that if the fastest electron does not make it to the collector plate, none of the others will, and there will be no photo current. Rearranging the photoelectric equation-

$\frac{1}{2}m{V}_{max}^2=hv-w\Rightarrow \frac{1}{2}m{\left(\frac{qB{r}_{max}}{m}\right)}^2=hv-w(rearrangingr=\frac{mv}{qB})\Rightarrow \frac{q^2{B}^2{r}_{max}^2}{2m}=hv-w\Rightarrow r_{max}^2=\left[\frac{2m(hv-w)}{q^2{B}^2}\right]\Rightarrow r_{max}=\frac{\sqrt{2m(hv-w)}}{qB}$

It should be clear, then, that if $r_{max}\le d$ , there will be no photo current. Or the largest value of $r_{max}$ for which there is no current is d = 10 cms.

$\therefore$ for no photo current

$B_{min}=\frac{\sqrt{2m(hv-w)}}{qd}=\frac{\sqrt{2m(\frac{hc}{\lambda }-w)}}{qD}$ . . . . . . (1)

But, wait we don’t have a source of light with a single unique wavelength, but a range of wavelengths 400 nm – 600 nm. The smaller wavelength naturally covering more energy than the longer ones, leading to faster electrons and a longer $r_{max}$ (we need to work with the longest $r_{max}$ to achieve $B_{min}$ ). We will therefore use, $\lambda$ = 400 nm in equation (1). Putting in all values,

$B_{min}=2.85\times {10}^{-5}T$