Single Choice

The energy of an electron having de-Broglie wavelength $$'\lambda'$$ is: [h = Planck's constant, m = mass of electron]

A$$\dfrac{h}{2m \lambda}$$
B$$\dfrac{h^2}{2m \lambda^2}$$
Correct Answer
C$$\dfrac{h^2}{2m^2 \lambda^2}$$
D$$\dfrac{h^2}{2m^2 \lambda}$$

Solution

$$\lambda=\dfrac{h}{mv}$$

$$\implies v=\dfrac{h}{m\lambda}$$

$$K=\dfrac{1}{2}mv^2=\dfrac{1}{2}m\left(\dfrac{h}{m\lambda}\right)^2$$

$$\implies K=\dfrac{h^2}{2m\lambda^2}$$

Hence, the answer is option-(B).

SIMILAR QUESTIONS

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The energy of a photon is $$3 \times 10^{-12}$$ ergs. Its wavelength (in nm) will be:

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Calculate the de Broglie wavelength for a beam of electron whose energy is 100 eV:

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If the de-Broglie wavelength of a particle of mass $$m$$ is $$100$$ times its velocity then its value in terms of its mass $$(m)$$ and Planck's constant $$(h)$$ is:

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The de Broglie wavelength of a particle of mass 1 gram and velocity $$100\ { ms }^{ -1 }$$ is:

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The de Broglie wavelength of an electron in the 4th Bohr orbit is:

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The de-Broglie's wavelength of electron present in first Bohr orbit of $$'H'$$ atom is?

Atomic Structure

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