(i) If the wave number of a beam of light is 400 cm-1 , then find out

1.	(i) If the wave number of a beam of light is 400 cm-1 , then find out its frequency. (ii) Write the de-Broglie's expression for dual nature of matter.
Answer» (i) Wave Number (\(\bar{v}\)) = 400 cm⁻¹ = 4 m⁻¹ \(\therefore\)wavelength \((\lambda)\) = \(\frac{1}{\bar v}\) = \(\frac{1}{4}\) = 0.25 m \(\because\) v = \(\frac{c}{\lambda}\) = \(\frac{3.0\times10^8ms^{-1}}{0.25m}\) = 12 × 10⁸s⁻¹ = 1.2 × 10⁹ s⁻¹ (ii) de-Broglie equation \(\lambda=\frac{h}{p}\) = \(\frac{h}{mv}\) Where \(\lambda\) = Wavelength of the particle m = mass of the particle p = momentum of the particle v = velocity of the particle h = Planck ′ s Constant

(i) If the wave number of a beam of light is 400 cm-1 , then find out its frequency. (ii) Write the de-Broglie's expression for dual nature of matter.

Answer»

(i) Wave Number (\(\bar{v}\)) = 400 cm⁻¹ = 4 m⁻¹

\(\therefore\)wavelength \((\lambda)\) = \(\frac{1}{\bar v}\) = \(\frac{1}{4}\) = 0.25 m

\(\because\) v = \(\frac{c}{\lambda}\) = \(\frac{3.0\times10^8ms^{-1}}{0.25m}\)

= 12 × 10⁸s⁻¹

= 1.2 × 10⁹ s⁻¹

(ii) de-Broglie equation

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

Where \(\lambda\) = Wavelength of the particle

m = mass of the particle

p = momentum of the particle

v = velocity of the particle

h = Planck ′ s Constant