In a photoelectric effect experiment, the slope of the graph between the stopping potential and the incident frequency will be ( textin SI textunit)

Question

In a photoelectric effect experiment, the slope of the graph between the stopping potential and the incident frequency will be ( textin SI textunit) (A) 10- 14 (B) 5× 10- 14 (C) 4× 10- 15 (D) 4× 10- 34

Tardigrade Admin · Accepted Answer

When light waves of differing frequencies are made incident on the same surface, the stopping potential (

V_{s}

) differ as is evidenced from Einstein's photoelectric equation given by

h v - h v_{0} = e V_{s}

where

h

is Planck's constant,

v, v_{0}

the frequencies and

e

the charge.
Solution

Hence,

V_{s} = \frac{h v}{e} - \frac{h v}{e}

...(i)
Comparing Eq. (i) with straight line equation

y = m x + c

Slope

m = \frac{h}{e}

Wh ere h = 6.6 \times 1 0^{- 34} J s, e = 1.6 \times 1 0^{- 19} C .

∴ m = \frac{6.6 \times 1 0 ^{- 34}}{1.6 \times 1 0 ^{- 19}} = 4.125 \times 1 0^{- 15}

Q. In a photoelectric effect experiment, the slope of the graph between the stopping potential and the incident frequency will be $(in S I unit)$

$1 0^{- 14}$

$5 \times 1 0^{- 14}$

$4 \times 1 0^{- 15}$

$4 \times 1 0^{- 34}$

Q. In a photoelectric effect experiment, the slope of the graph between the stopping potential and the incident frequency will be (inSIunit)

10−14

5×10−14

4×10−15

4×10−34

Q. In a photoelectric effect experiment, the slope of the graph between the stopping potential and the incident frequency will be $(in S I unit)$

$1 0^{- 14}$

$5 \times 1 0^{- 14}$

$4 \times 1 0^{- 15}$

$4 \times 1 0^{- 34}$