NEET Solved Paper 2018 Question 12
Question: An electron of mass m with an initial velocity $ \overrightarrow{{}v}=v _0\widehat{i}(v _0>0) $ enters an electric field $ \overrightarrow{{}E}\text{=-}{E_0}\widehat{i}\text{(}{E_0}\text{=} $ constant > 0) at t = 0. If $ {\lambda_0} $ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is [NEET - 2018]
Options:
A) $ {\lambda_0}t $
B) $ {\lambda_0}( 1+\frac{eE _0}{mV _0}t ) $
C) $ \frac{{\lambda_0}}{( 1+\frac{eE _0}{mV _0}t )} $
D) $ {\lambda_0} $
Show Answer
Answer:
Correct Answer: C
Solution:
Initial de-Broglie wavelength $ {\lambda_0}\text{=}\frac{h}{m{V_0}} $ …(i)
Acceleration of electron $ a=\frac{eE _0}{m} $
Velocity after time t, $ \text{V=}( {V_0}\text{+}\frac{e{E_0}}{m}t ) $
So, $ \lambda \text{=}\frac{h}{mV}\text{=}\frac{h}{m( {V_0}\text{+}\frac{e{E_0}}{m}t )} $ $ =\frac{h}{mV _0[ 1+\frac{eE _0}{mV _0}t ]}=\frac{{\lambda_0}}{[ 1+\frac{eE _0}{mV _0}t ]} $ …(ii)
Divide (ii) by (i), $ \lambda =\frac{{\lambda_0}}{[ 1+\frac{eE _0}{mV _0}t ]} $
BETA
