Question

The kinetic energy of the particle is$\mathrm{E}$. The de–Broglie wavelength is$\mathrm{\lambda }$. On increasing its K.E by$\mathrm{\Delta }$, its new de–Broglie wavelength becomes$\frac{\mathrm{\lambda }}{2}$. Then $\mathrm{\Delta }$ is

• A. $3E$
• B. $E$
• C. $2E$
• D. $4E$

Answer 1

The correct option is A

$3E$

Step 1. Given data:

1. The kinetic energy of the particle is$=\mathrm{E}$
2. The de–Broglie wavelength is$=\mathrm{\lambda }$
3. The kinetic energy is increased by$∆$
4. New de–Broglie wavelength$=\frac{\mathrm{\lambda }}{2}$

Step 2. Formula used:

1. The formula of de-Broglie wavelength is given by,
   $\mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}}$ [where $\mathrm{h}=$Planck's constant and $\mathrm{p}=$momentum of the particle]
2. The relation between kinetic energy and momentum is given by,
   $\mathrm{E}=\frac{{\mathrm{p}}^2}{2\mathrm{m}}$ [where $\mathrm{m}=$the mass of the particle]
3. The relation between kinetic energy and de-Broglie wavelength is given by,
   $\mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}\mathrm{as}\mathrm{p}=\sqrt{2\mathrm{mE}}$

Step 3. Calculating the value$∆$,

Let, the initial kinetic energy be$\mathrm{E}$.

final kinetic energy became$\left(\mathrm{E}+∆\right)$.

then, the ratio of initial momentum to final momentum is,

$$\begin{gathered} \frac{\mathrm{\lambda }}{\left({\displaystyle \frac{\mathrm{\lambda }}{2}}\right)}=\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}\times \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}+∆\right)}}{\mathrm{h}} \\ \Rightarrow 2=\frac{\sqrt{\left(\mathrm{E}+∆\right)}}{\sqrt{\mathrm{E}}} \\ \Rightarrow 4=\frac{\mathrm{E}+∆}{\mathrm{E}} \\ \Rightarrow ∆=3\mathrm{E} \end{gathered}$$

Thus, the value of$∆$ is$3\mathrm{E}$.

Hence, option A is the correct answer.