A stream of electrons from 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/λ (where λ is wavelength associated with electron wave) is given by

Question

A stream of electrons from 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/λ (where λ is wavelength associated with electron wave) is given by (A) √2 m e V (B) meV (C) 2meV. (D) √m e V

Tardigrade Admin · Accepted Answer

As electron of charge 'e' is passed through 'V' volt, kinetic energy of electron becomes ='eV' As wavelength of e- wave (.λ .)=(h/√2 m . K . E .) λ =(h/√2 m e V) So (h/λ )=√2 m e V

Q. A stream of electrons from 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 $h / λ$ (where $λ$ is wavelength associated with electron wave) is given by

$2 m e V$

meV

2meV.

$m e V$

As electron of charge 'e' is passed through 'V' volt, kinetic energy of electron becomes $=^{'} e V^{'}$
As wavelength of $e^{-}$ wave $(λ) = \frac{h}{2 m . K . E .}$
$λ = \frac{h}{2 m e V}$
So $\frac{h}{λ} = 2 m e V$

Q. A stream of electrons from 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/λ (where λ is wavelength associated with electron wave) is given by

2meV​

meV

2meV.

meV​

As electron of charge 'e' is passed through 'V' volt, kinetic energy of electron becomes =′eV′ As wavelength of e− wave (λ)=2m.K.E.​h​ λ=2meV​h​ So λh​=2meV​

Q. A stream of electrons from 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 $h / λ$ (where $λ$ is wavelength associated with electron wave) is given by

$2 m e V$

$m e V$

As electron of charge 'e' is passed through 'V' volt, kinetic energy of electron becomes $=^{'} e V^{'}$
As wavelength of $e^{-}$ wave $(λ) = \frac{h}{2 m . K . E .}$
$λ = \frac{h}{2 m e V}$
So $\frac{h}{λ} = 2 m e V$