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  • When a given photosensitive material is irradiated with light of frequency , the maximum speed of the emitted photoelectrons equals<img alt="V_{max}" src="https://ent

When a given photosensitive material is irradiated with light of frequency v, the maximum speed of the emitted photoelectrons equalsV_{max}. The graph shown in the figure gives a plot of  V^{2}_{max} varying with frequency v.

Obtain an expression for

(a) Planck’s constant, and

(b) The work function of the given photosensitive material in terms of the parameters ‘l’, ‘n’ and the mass ‘m’ of the electron.

(c) How is threshold frequency determined from the plot?

 

 
 

Answers (1)

(a) Einstein photoelectric equation can be used to determine the plank's constant 'h'. thus, we have

         K=\frac{1}{2}MV^{2}_{max}=h\nu -\phi_{0}

    \therefore V^{2}=\left ( \frac{2h}{m} \right )\nu -\frac{2\phi _{0}}{m}                \left [ \because \phi =h\nu _{0} \right ]

We know, the slope of the graph is

 \frac{2h}{m}=\frac{l}{n}

 plank's constant 

 h=\frac{ml}{2n}

(b) From the graph

        \frac{2\phi _{0}}{m}=l  (intercept on the vertical axis)

Hence, the work function will be;-

        \phi _{0}=\frac{ml}{2}        

(c) The threshold frequency is the intercept on horizontal

Therefore, \nu _{0}=n

Posted by

Safeer PP

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