Question

The debroglie wave length of an electron in a certain orbit of hydrogen atom is 13.3. then
the no. of waves present in that orbit is​

Answer 1

Answer:

The de Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is 2π×0.529Ao.

First Bohr orbit of 'H' atom has radius r=0.529Ao

Also, the angular momentum is quantised.

mvr=2πh​

2πr=mvh​=λ

λ=2π×0.529Ao

So, the correct option is B

Explanation:

Answer 2

Concept:

• DeBroglie wavelength
• De Broglie wavelength is the wavelength () that is connected to an item in relation to its momentum and mass. Typically, a particle's force is inversely proportional to its de Broglie wavelength.
• Quantum mechanics
• Calculating wave number
• The number of wavelengths per unit distance of the spatial wave frequency, also known as the spatial frequency or angular wavenumber in physics, is known as wave number. It's a scalar amount.

Given:

• DeBroglie wavelength λ = 13.3 nm = 1.33A
• Hydrogen atom
• Radius of atom = 0.529A

Find:

• The number of waves

Solution:

The number of waves = circumference of orbit/wavelength

n = 2πR/λ

n = 2*3.14*0.529/1.33

n = 2.5

The number of waves present is 2.5.

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