Question

A small piece of Cesium metal having work function $(\varphi =1.9\text{eV})$ is kept at a distance of $20\text{cm}$ from a large metal plate having a charge density of $1.0\times {10}^{-9}{\text{C m}}^{-2}$ on the surface facing the small Cesium piece. A monochromatic light of wavelength $400\text{nm}$ is incident on the Cesium piece. Find the minimum and the maximum kinetic energy of the photoelectrons reaching the large metal plate. Neglect any change in electric field due to the small piece of cesium present.$(hc=1240\text{eV nm})$

• A. $22.6\text{eV};23.8\text{eV}$
• B. $0\text{eV};11.3\text{eV}$
• C. $32\text{eV};48\text{eV}$
• D. $9\text{eV};10\text{eV}$

Answer 1

The correct option is A $22.6\text{eV};23.8\text{eV}$

Given,

$\sigma =1\times {10}^{-9}{\text{C m}}^{-2};{\varphi }_{Cs}=1.9\text{eV}d=20\text{cm};\lambda =400\text{nm}$

The electric potential difference between the charged plates is,

$V=E\times d=\frac{\sigma }{{\epsilon }_0}\times d[\because E=\frac{\sigma }{{\epsilon }_0}]$

$=\frac{1\times {10}^{-9}\times 20\times {10}^{-2}}{8.85\times {10}^{-12}}\approx 22.6\text{V}$

From, Einstein's photoelectric equation,

$e{V}_0=E-{\varphi }_0$

$e{V}_0=\frac{hc}{\lambda }-{\varphi }_0=\frac{1240}{400}-1.9$

$V_0=3.1-1.9=1.2\text{V}$

As $V_0$ is much less than $V$

Hence, the minimum energy required to reach the charged plate will be,

$K{E}_{min}=e{V}_0=22.6\text{eV}$

KE of the emittted electron from the ceisium plate lies from zero to $K{E}_{max}$ but the pd between the plates will provide the extra energy.

Hence, $K{E}_{max}=e{V}_0+eV=1.2+22.6=23.8\text{eV}$

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