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If we plot a graph of radiation against the stopping potential of photoelectron, its slope would be equal to

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Q. If we plot a graph of stopping potential of photoelectrons against the frequency (hertz) of radiation that has emitted it, then it's slope would be:

Q. If we plot a graph of stopping potential of photoelectrons against the frequency (in hertz) of radiation that has emitted, then its slope would be:
(

h = 6.6 \times 10^{- 34})

Q. The maximum kinetic energy (

E_{k}

) of emitted photoelectrons against frequency

ν

of incident radiation is plotted as shown in fig. The slope of the graph is equal to

Q. Find the slope of the graph shown in the figure.

V_{s} \to

stopping potential

(V)

f \to

frequency of the incident radiation

(Hz)

Q. When radiation is incident on a photoelectron emitter, the stopping potential is found to be

9 V

e / m

for the electron is

1.8 \times 10^{11} C k g^{- 1}

. What is the max. speed of the photoelectrons?

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