Q.463. de Broglie wavelength of electron in hydrogen atom

Question 463.
What is the wavelength of electron in n'th Bohr orbit ?

Answer 463.
For the electron of hydrogen atom,
mv^2/r = ke^2/r^2
=> mv^2 * r = ke^2

ByBohr's first hypothesis,
mvr = nh/2π

Taking ratio,
v = 2πke^2/nh
=> mv = 2mπ * ke^2/nh

de Broglie wavelength of the elctron
= h/mv
= nh^2 / 2 π mke^2
= 2 π n * [(1/mk) * (h/2π*e)^2]
= 2 π n * [1/(9.1 x 10^-31 x 9 x 10^9) * (6.62 x 10^-34/2π x 1.6 x 10^-19)^2 m
= 2 π n * 0.00529 x 10^8 m
= 2 π n * 0.529 angstrom units ... [1 angstrom = 10^-10 m].

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Q.341. de Broglie wavelength.

Question 341.
Calculate the de Broglie wavelength for a proton with a kinetic energy of 100 eV.

Answer 341.
E = (1/2) mv^2 = (mv)^2 / (2m = p^2/2m
=> p = √(2mE)

de Broglie wavelength,
λ = h/p


Mass of proton = 1.67262158 × 10-27 kilograms
Planck's constant, h = 6.626068 × 10-34 m2 kg / s
K.E. given = 100 eV = 100 x 1.60217646 × 10-19 joule

=> de Broglie wavelength, λ
= (6.626068 × 10-34) / [√(2 * 1.67262158 x 10^-27) * (100 x 1.60217646 x 10^-19)) m
= 2.862 x 10-12 m.

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