Question

A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of $$h /\lambda$$ (where $$\lambda$$ is wavelength associated with electron wave) is given by :

A
2meV
B
meV
C
2meV
D
meV

Solution

The correct option is C $$\sqrt{2meV}$$Kinetic energy $$\displaystyle =eV = \dfrac {1}{2} mu^2$$$$\displaystyle u = \sqrt {\dfrac {2eV}{m}}$$......(1)Here, u is the velocity of the electron.The de broglie wavelength is $$\displaystyle \lambda = \dfrac {h}{mu}$$.$$\displaystyle \dfrac {h}{\lambda} = mu$$......(2)Substitute equation (1) in equation (2)$$\displaystyle \dfrac {h}{\lambda} = m \sqrt {\dfrac {2eV}{m}} = \sqrt {2meV}$$Hence, the correct option is $$C$$Chemistry

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