In an experiment on photoelectric effect the frequency $v$ of the incident light is plotted against the stopping potential $V_{0}$ . The work function of the photoelectric surface is given by (e is electron charge)

O B \times e i n e V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

O B i n v o l t

No worries! We‘ve got your back. Try BYJU‘S free classes today!

OA in

e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

T h e s l o p e o f t h e l i n e A B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $O B \times e i n e V$
Using Einstein's equation ,
$V_{0} = \frac{h}{e} ν - \frac{W_{0}}{e}$

Comparing this equation with $y = m x + c$ ,
we get on $(- V_{0})$ axis,

$O B = \frac{W_{0}}{e}$

hence, $W_{0} = O B \times e$ eV.

Suggest Corrections

Similar questions

Q. In a photoelectric experiment, the stopping potential

V_{s}

is plotted against the frequency

v

of incident light. The resulting curve is a straight line which makes an angle

θ

which is equal to (

φ

= work function of surface)

Figure (42-E2) is the plot of the stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find (a) the ratio h/e and (b) the work function.

Q. The figure is the plot of stopping potential versus the frequency of the light used in an experiment on photoelectric effect. Find (a) the ratio h/e and (b) the work function.

Q. In a photoelectric experiment, the graph of frequency

υ

of incident light (in Hz) and stopping potential V (in V) is as shown in the figure. From figure, the value of the Plancks constant is (e is the elementary charge)

Q. For photoelectric effect in a metal, the graph of the stopping potential

V_{0}

(in volt) versus frequency

v

(in hertz) of the incident radiation is shown in Fig. The work function of the metal (in

e V

) is

Join BYJU'S Learning Program

In an experiment on photoelectric effect the frequency v of the incident light is plotted against the stopping potential V0. The work function of the photoelectric surface is given by (e is electron charge)

The correct option is A OB×e in eVUsing Einstein's equation ,V0=heν−W0eComparing this equation with y=mx+c,we get on (−V0) axis,OB=W0ehence, W0=OB×e eV.

In an experiment on photoelectric effect the frequency $v$ of the incident light is plotted against the stopping potential $V_{0}$ . The work function of the photoelectric surface is given by (e is electron charge)

The correct option is A $O B \times e i n e V$
Using Einstein's equation ,
$V_{0} = \frac{h}{e} ν - \frac{W_{0}}{e}$

Comparing this equation with $y = m x + c$ ,
we get on $(- V_{0})$ axis,

$O B = \frac{W_{0}}{e}$

hence, $W_{0} = O B \times e$ eV.