de Broglie wavelength of an electron after being accelerated by a potential difference of V volt from rest is :
A
`lamda=(1.23)/sqrt(m)`
B
`lamda=(1.23)/sqrt(h)m`
C
`lamda=(1.23)/sqrt(V)nm`
D
`lamda=(1.23)/(V)`
Text Solution
Verified by Experts
Topper's Solved these Questions
ATOMIC STUCTURE
NARENDRA AWASTHI|Exercise Level- 1|1 Videos
ATOMIC STUCTURE
NARENDRA AWASTHI|Exercise level 2|1 Videos
CHEMICAL EQUILIBRIUM
NARENDRA AWASTHI|Exercise Level 2|1 Videos
Similar Questions
Explore conceptually related problems
What is de -Brogile wavelength of an eectron beam accelerated through a potential difference of 36 V?
What is the de-Broglie wavelength of an electron beam accelerated through a potential difference of 25 V?
Calculate the momentum and De Brogile wavelength of the electrons accelerated through a potential difference of 56V. Mass of electron is 9.1 xx 10^(-31) kg.
Calculate the momentum and de-Broglie wavelength of the electrons accelerated through a potential difference of 56V.
Derive the expression for de Broglie wavelength associated with an electron in a potential differences of V volts.