JEE MAIN - Physics (2025 - 29th January Morning Shift - No. 1)
If $\lambda$ and $K$ are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be :
_29th_January_Morning_Shift_en_1_1.png)
_29th_January_Morning_Shift_en_1_2.png)
_29th_January_Morning_Shift_en_1_3.png)
_29th_January_Morning_Shift_en_1_4.png)
Explanation
$$\begin{aligned}
&\begin{aligned}
& \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}} \\
& \lambda^2=\frac{\mathrm{h}^2}{2 \mathrm{~m}}\left(\frac{1}{\mathrm{k}}\right) \\
& \mathrm{Y}=\mathrm{cx}^2
\end{aligned}\\
&\text { Upward facing parabola passing through origin. }
\end{aligned}$$
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