1.

For an alpha particle, accelerated through a potential difference V, wavelength (in Å) of the associated matter wave is1. \(\frac{12.27}{\sqrt{V}}\)2. \(\frac{0.101}{\sqrt{V}}\)3. \(\frac{0.202}{\sqrt{V}}\)4. \(\frac{0.286}{\sqrt{V}}\)

Answer» Correct Answer - Option 2 : \(\frac{0.101}{\sqrt{V}}\)

CONCEPT :

  • De Broglie wavelength connects between the wavelength and momentum of the particle and is given by, and is given by

\(⇒ λ = \frac{h}{P} = \frac{h}{mV}\)----(1)

Where m = mass of the particle and V = Velocity, h = Plancks constant

Substituting the value\(P = \sqrt{2mE}\)in equation 1, it can be rewritten as

\(⇒ λ = \frac{h}{\sqrt{2mE}}\)

Where m = mass, E= kinetic energy,

EXPLANATION :

  • The de Broglie wavelength is given by

\(⇒ \lambda =\frac{h}{P}\)

or\(λ=\frac{h}{\sqrt{2mE}}\)--------(1)

  • The kinetic energy of particle accelerated through potential V is given by

⇒ E = q × V

Where q = Charge, V = potential difference

For an alpha particle q = 2e

⇒E = 2e × V

and M = 4mP

Substituting the given values in equation 1

\(⇒ λ=\frac{h}{\sqrt{4m_P × 2e× V}}\)

Substituting the given values mp= 1.67 × 10-27Kg, e = 1.6 × 10-19C, h = 6.626 × 10-34Joues/sec and solving we get

\(⇒ λ=\frac{0.101}{\sqrt{V}}\)

  • Hence, option 2 is the answer


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