CBSE 2024 · Region 3 · Set 2 · Q20 · 2 marks
Light of wavelength $\displaystyle 500$ nm is incident on caesium metal (work function $\displaystyle 2 \cdot 14 \mathrm{eV}$ ) and photoemission of electrons occurs. Calculate the (i) kinetic energy (in eV) of the fastest electrons and (ii) stopping potential for this situation. (Take hc $\displaystyle =1240 \mathrm{eV} . \mathrm{nm}$ )
Marking-scheme solution
Using Einstein Photoelectric equation
\[\begin{aligned}
\frac{h c}{\lambda} & =K \cdot E_{\max }+\phi_{0} \\
K \cdot E_{\max } & =\frac{h c}{\lambda}-\phi_{0} \\
& =\frac{1240 \mathrm{eVnm}}{500 \mathrm{~nm}}-2.14 \mathrm{eV}
\end{aligned}
\]
\[\begin{aligned}
& K . E_{\max }=0.34 e V \\
& K . E_{\max }=e V_{0}
\end{aligned}
\]
\[\therefore V_{0}=0.34 V
\]
Dual Nature of Radiation and MatterEinstein’s Photoelectric Equation: Energy Quantum of RadiationApplyvery_short_answermedium
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.