Single Choice

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

A2πa0
B4πa0
C8πa0
Correct Answer
D6πa0

Solution

De-Broglie wavelength of an electron:

$$2\pi r = n\lambda$$

$$2\pi \times \dfrac{n^2}{z} a_0 = n.\lambda$$

$$2\pi \times \dfrac{4^2}{1}a_0 = 4\lambda$$

$$\lambda = 8\pi a_0$$

Hence, the correct answer is option $$C$$.

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:

Atomic Structure

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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Threshold wavelength of a metal is $${\lambda}_{0}$$. The de Broglie wavelength of photoelectron when the metal is irradiated with the radiation of wavelength $$\lambda$$ is:

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The de Broglie wavelength associated with a ball of mass 200 g and moving at a speed of 5 meters / hour, is of the order of ($$h= 6.625\times 10^{-34}J s $$) 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 (λ) associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v0 is thrshold frequency]:

Atomic Structure

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

Atomic Structure

Photoelectrons are liberated by ultraviolet light of wavelength $$3000\mathring{A}$$ from a metallic surface for which the photoelectric threshold is $$4000\mathring{A}$$. The de-Broglie wavelength of electrons emitted with maximum kinetic energy is:

Atomic Structure

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

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