Question

In an experiment on photoelectric effect, the emitter and the collector plates are placed at a separation of $10\text{cm}$ 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 $2.39\text{eV}$ and the light incident on it has wavelengths between $400\text{nm}$ and $600\text{nm}$. Find the minimum value of $B$ for which the current registered by the ammeter is zero. Neglect any effect of space charge.

• A. $5.87\times {10}^{-5}\text{T}$
• B. $1.42\times {10}^{-5}\text{T}$
• C. $2.85\times {10}^{-5}\text{T}$
• D. $8.73\times {10}^{-5}\text{T}$

Answer 1

The correct option is C $2.85\times {10}^{-5}\text{T}$

Given,
${\varphi }_0=2.39\text{eV};d=10\text{cm}{\lambda }_1=400\text{nm};{\lambda }_2=600\text{nm}$


For photo current to be zero, the emitted photoelectrons should not reach the collector. Due to the presence of the magnetic field $\left(B\right)$ the emitted photoelectrons will undergo a circular motion of radii (r) equal to the separation between the plates.

For minimum value of $\left(B\right)$, energy of the photoelectrons should be maximum, Thus $\lambda$ should be minimum.

Now, energy of the emitted photoelectrons is,

$E=\frac{hc}{\lambda }-{\varphi }_0=\frac{1240}{400}-{\varphi }_0$

$3.1-2.39=0.71\text{eV}$.

The radius of the circular path is,

$r=\frac{mv}{qB}$

$\Rightarrow r=\frac{\sqrt{2mE}}{qB}$

$\Rightarrow 0.1=\frac{\sqrt{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times 0.71}}{1.6\times {10}^{-19}\times B}$

$\Rightarrow B=\sqrt{\frac{12.922\times {10}^{-12}}{0.016}}$

$\Rightarrow B\approx 2.85\times {10}^{-5}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.

Why this question? Key concept: This type of multiconcepts question have been asked many times in JEE Advance.