Surface of certain metal is first illuminated with light of wavelength

Question:

Surface of certain metal is first illuminated with light of wavelength $\lambda_{1}=350 \mathrm{~nm}$ and then, by light of wavelength $\lambda_{2}=540 \mathrm{~nm}$. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of (2) The work function of the metal (in $\mathrm{eV}$ ) is close to:

(Energy of photon $\left.=\frac{1240}{\lambda(\text { in } \mathrm{nm})} \mathrm{eV}\right)$

  1. (1) $1.8$

  2. (2) $2.5$

  3. (3) $5.6$

  4. (4) $1.4$


Correct Option: 1

Solution:

(1) From Einstein's photoelectric equation,

$\frac{\mathrm{hc}}{\lambda_{1}}=\phi+\frac{1}{2} \mathrm{~m}(2 \mathrm{v})^{2}$  ...(1)

and $\frac{\mathrm{hc}}{\lambda_{2}}=\phi+\frac{1}{2} \mathrm{mv}^{2}$                  ...(2)

As per question, maximum speed of photoelectrons in two cases differ by a factor 2

From eqn. (i) \& (ii)

$\Rightarrow \frac{\frac{h c}{\lambda_{1}}-\phi}{\frac{h c}{\lambda_{2}}-\phi}=4 \Rightarrow \frac{h c}{\lambda_{1}}-\phi=\frac{4 h c}{\lambda_{2}}-4 \phi$

$\Rightarrow \frac{4 \mathrm{hc}}{\lambda_{2}}-\frac{\mathrm{hc}}{\lambda_{1}}=3 \phi \Rightarrow \phi=\frac{1}{3} \mathrm{hc}\left(\frac{4}{\lambda_{2}}-\frac{1}{\lambda_{1}}\right)$

$=\frac{1}{3} \times 1240\left(\frac{4 \times 350-540}{350 \times 540}\right)=1.8 \mathrm{eV}$

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