Question

De Broglie wavelength of an electron after being accelerated by a potential difference of V volt from rest is

• A. $\lambda = \frac { 12.3 } { \sqrt { \mathrm {~h} } } \AA$
• B.
• C.
• D. $\lambda = \frac { 12.3 } { \sqrt { \mathrm {~m} } } \AA$

Answer 1

Answer: (B)

Answer 2

For an electron accelerated with potential difference V volt, $\lambda = \frac { \mathrm { h } } { \sqrt { 2 \mathrm { mqV } } } = \frac { 12.3 } { \sqrt { \mathrm {~V} } } \AA$.