Question

In photoelectric effect, initially when energy of
electrons emitted is Eo, de-Broglie wavelength
associated with them is lo. Now, energy is doubled
then associated de-Broglie wavelength ' is
​

Answer 1

Answer:

de-Brogile wavelength is given by

λ=

p

h

, where h= Planck's constant and p= momentum

Also energy (E) and momentum are related as

E=

2m

p

2

∴p=

2mE

∴λ=

2mE

h

×

E

1

as h and m are constants

Hence,

λ

′

λ

0

=

E

E

′

=

E

2E

=

2

∴λ

′

=

2

λ

0