Question

de-broglie wavelength of electron in the fifth shell of hydrogen atom will be?

Answer 1

Answer:

Explanation:

For an electron with KE = 1 eV and rest mass energy 0.511 MeV, the associated DeBroglie wavelength is 1.23 nm, about a thousand times smaller than a 1 eV photon. (This is why the limiting resolution of an electron microscope is much higher than that of an optical microscope.)

Answer 2

De-broglie wavelength of electron in the fifth shell of hydrogen atom will be 16.61 Å.

Explanation:

Radius of fifth orbit $=r=R \times(5)^{2}$

= 25 R

Here R is the radius of the first orbit of hydrogen atom.

According to the law of conservation of angular momentum

$m \times v \times r=\frac{n h}{2 \pi}$

Or $\frac{h}{m v}=\frac{2 \pi r}{n}$

Here n is equal to 5

$\lambda=\frac{h}{m v}=2 \pi \times \frac{25 R}{5}$

Or $\lambda=10 \pi R$

Or $\lambda=10 \pi \times(0.529 Å)$, since the value of r is equal to 0.529 angstrom

Or λ = 16.61 Å

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