A particle having electric charge 3 10⁻¹⁹ C and mass 6 10⁻²⁷ ~kg is accelerated by applying an electric potential of 1.21 V. Wavelength of the matter wave associated with the particle is 10⁻¹² ~m . The value of is _ _ _ _ - (Take Planck's constant =6.6 10⁻³⁴ ~J . s )

Step-by-step solution

The de Broglie wavelength is = h mv where v is found from energy conservation. From qV = 1 2 mv^2 : v = 2qV m = 2 3 10⁻¹⁹ 1.21 6 10⁻²⁷ = 1.21 10^8 = 1.1 10^4 m/s. Therefore: = 6.6 10⁻³⁴ 6 10⁻²⁷ 1.1 10^4 = 6.6 10⁻³⁴ 6.6 10⁻²³ = 10⁻¹¹ m = 10 10⁻¹² m. Thus = 10 .