Question

Maximum velocity of photo electrons
emitted by a photo meter is 1.8 x 10m/s.
Taking in 1.8 x 1011 C/kg for electrons,
the stopping potential of emitter is​

Answer 1

Answer:

The stopping potential of emitter is 9V.

K.E.

max

=(hγ−ϕ)

2

1

mv

2

=hγ−ϕ ----------------(1)

stopping potential=(hγ−ϕ)/e

V=(hγ−ϕ)/e ----------------(2)

Now, putting value of (hγ−ϕ) in (2) from (1),

V=(

2

1

mv

2

)/e

=

2e

mv

2

=

2e/m

v

2

=

2×1.8×10

11

1.8×10

6

××1.8×10

6

=9V.