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In the experiment on the photoelectric effect, the graph between E\nu (max) and \nu is found to be a straight line as shown in fig


The threshold frequency and Planck's constant according to this graph are 

Option: 1

3.33\times 10^{+18}, 6\times 10^{-34 }J-s

 

 

 


Option: 2

6\times 10^{18}s^{-1}, 6\times 10^{-34 }J-s


Option: 3

2.66\times 10^{18}s^{-1}, 4\times 10^{-34 }J-s


Option: 4

4\times 10^{18}s^{-1}, 3\times 10^{-34 }J-s


Answers (1)

best_answer

As we learned

 

Threshold frequency -

The minimum frequency of incident radiation to eject electron is threshold frequency (\nu_{0})

- wherein

If incident frequency v< \nu _{0}

No photoelectron emission

 

 

E_{max }=h\nu -\phi

\phi is intercept on y-axis and h is the slope.

\therefore h=\frac{2.4\times 10^{-15}}{4\times 10^{18}}=6\times 10^{-34}J-s

\phi =2\times 10^{-15}J \Rightarrow h\nu _{0}=2\times 10^{-15}

\nu _{0}=3.33\times 10^{+18}

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manish

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