CBSE 2023 · Region 3 · Set 2 · Q26 · 3 marks
(a)A particle of mass $\displaystyle m$ and charge $\displaystyle q$ is accelerated through a potential difference V. Plot a graph of de Broglie wavelength $\displaystyle \lambda$ associated with it as a function of $\displaystyle V$.(b)Calculate the energy acquired by and de Broglie wavelength associated with, an electron accelerated through a potential difference of $\displaystyle 400$ V .
(a)
A particle of mass $\displaystyle m$ and charge $\displaystyle q$ is accelerated through a potential difference V. Plot a graph of de Broglie wavelength $\displaystyle \lambda$ associated with it as a function of $\displaystyle V$.
(b)
Calculate the energy acquired by and de Broglie wavelength associated with, an electron accelerated through a potential difference of $\displaystyle 400$ V .
Marking-scheme solution
(a) Graph
(b) Calculation of energy & de-Broglie Wavelength $\displaystyle 1$+ $\displaystyle 1$
(a) 
(b) Energy acquired \(\displaystyle =q V_{o}\)
\[=1.6 \times 10^{-19} \times 400
\]
\[=6.4 \times 10^{-17} \mathrm{~J}=400 \mathrm{eV}
\]
\[\lambda=\frac{h}{\sqrt{2 m K}}
\]
\[\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 6.4 \times 10^{-17}}}
\]
\[\lambda=0.6 \mathrm{k} 10^{-10} m=0.6145 \mathrm{~A} 0
\]Dual Nature of Radiation and MatterWave Nature of MatterApplyshort_answermedium
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CBSE Class 12 Physics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.