de-Broglie wavelength associated with an electron accelerated through a potential difference V is `lambda`. What will be its wavelength when the accelerating potential is increased to 4V?

Options

A. `lamda`
B. `lamda/2`
C. `2lamda`
D. `1.5lamda`

Answer

B

Explanation

The de broglie wavelength `lamda = (h)/(sqrt(2meV))`
where V is the P.D. Thus `lamda prop (1)/(sqrtV)`
`therefore lamda_2/lamda_1 = sqrt(V_1/V_2) = sqrt((V)/(4V)) = 1/2 therefore lamda_2 = (lamda_1)/(2) " or " lamda_2`