1.

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

Answer»

increased by `(lambda)/(2)`
decrease by `(lambda)/(2)`
increased by 2`lambda`
decreased by `2lambda`

SOLUTION :`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)`.


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