1.

When a certain metal was irradiated with light of frequency3.2 x 1016 s-1, the photoelectrons emitted had twice thekinetic energy as did the photoelectrons emitted whenthe same metal was irradiated with light of frequency2.0x1016 s-1. Calculate threshold frequency for the metal.[Hint. K.E. = hv -hvo; hví - hvo = 2 (hv 2 - hvo);or Vo = 2v2 - Vi]​

Answer»

Answer:

Applying photoelectric EQUATION,

KE=hv−hv

0

or (v−v

0

)=

h

KE

Given, KE

2

=2KE

1

v

2

−v

0

=

h

KE

2

...(i)

and v

1

−v

0

=

h

KE

1

...(II)

Dividing equation (i) by equation (ii),

v

1

−v

0

v

2

−v

0

=

KE

1

KE

2

=

KE

1

2KE

1

=2

or v

2

−v

0

=2V

1

−2v

0

or v

0

=2v

1

−v

2

=2(2.0×10

16

)−(3.2×10

16

)

=8.0×10

15

Hz.


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