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Model Paper 2 — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryModel Paper 23 Marks
Question
A photon of wavelength $4 \times 10^{-7} m$ strikes on the metal surface, the work function of metal being 2.13 eV . Calculate
i. the energy of the photon $( eV )$,
ii. the kinetic energy of emission,
iii. the velocity of the photoelectron. $\left(1 eV =1.6020 \times 10^{-19} J\right)$.
✓

Answer

i. The energy of the photon
$
\begin{aligned}
& \text { Energy }(E)=\frac{hc}{\lambda}-\frac{\left(6.626 \times 10^{-34} Js\right) \times\left(3 \times 10^8 ms^{-1}\right)}{\left(4 \times 10^{-7} m\right)}=4.97 \times 10^{-19} \\
& =\frac{(1 eV)}{\left(1.602 \times 10^{-19} J\right)} \times\left(4.97 \times 10^{-19} J\right)=3.1 eV
\end{aligned}
$
ii. The kinetic energy of emission
Kinetic energy of emission $= E$ - work function (i.e. kinetic energy of emitted electron) $=(3.1-2.13)=0.97 eV$
iii. Velocity of photoelectron
$
\begin{aligned}
& \text { KE of emission }=\frac{1}{2} mv^2=0.97 eV \\
& =0.97 \times 1.602 \times 10^{-19} J=0.97 \times 1.602 \times 10^{-19} kg m^2 s^{-2} \\
& \text { or } v^2=\frac{2 \times 0.97 \times 1.602 \times 10^{-19}\left(kgm^2 s^{-2}\right)}{\left(9.1 \times 10^{-31} kg\right)}=0.34 \times 10^{12} m^2 s^{-2} \\
& \text { or } v=\left(0.34 \times 10^{12} m^2 s^{-2}\right)^{1 / 2}=0.583 \times 10^6 ms^{-1}=5.83 \times 10^5 ms^{-1}
\end{aligned}
$

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