(i) What is de-Broglie wavelength for an election with light velocity?

1.	(i) What is de-Broglie wavelength for an election with light velocity? (ii) What is angular momentum of electron in 5th shell?
Answer» (i) Mass of electron (m) = 9.1 × 10⁻³¹ Kg Velocity of light (v) = 3.0 × 10⁸ m/sec \(\therefore\) According to de-Broglie equation, \(\lambda\) = \(\frac{h}{mv}\) = \(\frac{6.626\times10^{-34}Js}{9.1\times10^{-31}kg\times3.0\times10^8m/s}\) (ii) Angular momentum of electron is 5^th shell J = \(\frac{nh}{2\pi}\) = \(\frac{5\times6.626\times10^{-34}Jsec}{2\times3.14}\) = \(\frac{33.13\times10^{-34}}{6.28}\) = 5.275 × 10⁻³⁴ kg m² /sec

(i) What is de-Broglie wavelength for an election with light velocity? (ii) What is angular momentum of electron in 5th shell?

Answer»

(i) Mass of electron (m) = 9.1 × 10⁻³¹ Kg

Velocity of light (v) = 3.0 × 10⁸ m/sec

\(\therefore\) According to de-Broglie equation,

\(\lambda\) = \(\frac{h}{mv}\)

= \(\frac{6.626\times10^{-34}Js}{9.1\times10^{-31}kg\times3.0\times10^8m/s}\)

(ii) Angular momentum of electron is 5^th shell

J = \(\frac{nh}{2\pi}\)

= \(\frac{5\times6.626\times10^{-34}Jsec}{2\times3.14}\)

= \(\frac{33.13\times10^{-34}}{6.28}\)

= 5.275 × 10⁻³⁴ kg m² /sec