The energy that should be added to an electron to reduce its de-Broglie wavelength from λ to 2 λ is n times the initial energy. The value of ' n ' is

Question

The energy that should be added to an electron to reduce its de-Broglie wavelength from λ to 2 λ ​ is n times the initial energy. The value of ' n ' is

TestHub · Accepted Answer

The de-Broglie wavelength of an electron is given by $λ = \frac{h}{2 m E}$ , where $E$ is the kinetic energy.

Squaring both sides, we get $E = \frac{h ^{2}}{2 m λ ^{2}}$ .

Let the initial energy be $E_{1} = \frac{h ^{2}}{2 m λ ^{2}}$ .

When the wavelength is reduced to $\frac{λ}{2}$ , the final energy becomes $E_{2} = \frac{h ^{2}}{2 m ( \frac{λ}{2} ) ^{2}} = 4 \frac{h ^{2}}{2 m λ ^{2}} = 4 E_{1}$ .

The energy that should be added is $Δ E = E_{2} - E_{1} = 4 E_{1} - E_{1} = 3 E_{1}$ .

Given that the added energy is $n$ times the initial energy, we have $n E_{1} = 3 E_{1}$ .

Thus, $n = 3$ .

Answer: 3

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