If speed of electron emitted becomes double change in wavelength will be …….

Options

A. increased by `(lambda)/(2)`
B. decrease by `(lambda)/(2)`
C. increased by 2`lambda`
D. decreased by `2lambda`

Answer

B

Explanation

`v_(2)=2v_(1)`
Kinetic energy of electron `E_(1)=(1)/(2)mv_(1)^(2)`
`E_(2)=(1)/(2)mv_(2)^(2)=(1)/(2)m(2v_(1))^(2)=(1)/(2)m(4v_(1)^(2))`
`therefore E_(2)=4[(1)/(2)mv_(1)^(2)] therefore E_(2)=4E_(1)`
Now `lambda=(h)/(sqrt(2mE))`
`therefore lambda prop (1)/(sqrt(E))`
`therefore (lambda_(2))/(lambda_(1))=sqrt((E_(1))/(E_(2)))=sqrt((E_(1))/(4E_(1)))=(1)/(2)therefore lambda_(2)=(lambda_(1))/(2)`
`therefore` Decrease in wavelength `lambda_(1)-(lambda_(1))/(2)=(lambda_(1))/(2)`.