Modern Physics Ques 106
- Surface of certain metal is first illuminated with light of wavelength $\lambda _1=350 $ $nm$ and then by light of wavelength $\lambda _2=540 $ $n-m$. It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of $2$ . The work function of the metal (in eV) is close to (energy of photon $=\frac{1240}{\lambda(\text { in } n-m)} eV$ )
(a) $5.6$
(b) $2.5$
(c) $1.8$
(d) $1.4$
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Answer:
Correct Answer: 106.(c)
Solution:
Formula:
- Let maximum speed of photo electrons in first case is $v _1$ and maximum speed of photo electrons in second case is $v _2$
Assumption I if we assume difference in maximum speed in two cases is $2$ then $v _1=v$ and $v _2=3 v$
According to Einstein’s photo electron equation Energy of incident photon $=$ work function $+KE$
$ \text { i.e. } \quad \frac{h c}{\lambda}=\phi _0+\frac{1}{2} m v^{2} $
Where $h c=1240 eV, \lambda$ is wavelength of light incident, $\phi _0$ is work function and $v$ is speed of photo electrons.
$ \begin{aligned} & \text { When } & \lambda _1 & =350 nm \\ & \therefore & \frac{h c}{350} & =\phi _0+\frac{1}{2} m v^{2} \\ & \text { or } & \frac{h c}{350}-\phi _0 & =\frac{1}{2} m v^{2} \quad …….(i)\\ & \text { when } & \lambda _2 & =540 nm \\ & \therefore & \frac{h c}{540} & =\phi _0+\frac{1}{2} m\left(3 v^{2}\right) \\ & \therefore & \frac{h c}{540}-\phi _0 & =\left(\frac{1}{2} m v^{2}\right) \times 9 \quad …….(ii) \end{aligned} $
Now, we divide Eq. (i) by Eq. (ii), we get
$\frac{\frac{h c}{350}-\phi _0}{\frac{h c}{540}-\phi _0} =\frac{\frac{1}{2} m v^{2}}{\left(\frac{1}{2} m v^{2}\right) \times 9}=\frac{1}{9} $
$\text { or } 9(\frac{h c}{350}-\phi _0) =\frac{h c}{540}-\phi _0 $
$\text { or } \quad 8 \phi _0 =h c [\frac{9}{350}-\frac{1}{540}] $
$\text { or } \phi _0 =\frac{1}{8} \times 1240 [\frac{9 \times 540-350}{350 \times 540}]=3.7$ $ eV .$
No option given is correct.
Alternate Method
Assumption II If we assume velocity of one is twice in factor with second, then.
Let $v _1=2 v$ and $v _2=v$
We know that from Einstein’s photoelectric equation, energy of incident radiation $=$ work function $+KE$
or $\quad \frac{h c}{\lambda}=\phi+\frac{1}{2} m v^{2}$
Let when $\lambda _1=350 $ $nm$ then $v _1=2 v$
and when $\lambda _1=540$ $ nm$ then $v _2=v$
$\therefore$ Above Eq. becomes
$ \begin{aligned} &\frac{h c}{\lambda _1} =\phi+\frac{1}{2} m\left(2 v^{2}\right) \\ \text { or } & \frac{h c}{\lambda _1}-\phi =\frac{1}{2} m \times 4 v^{2} \quad …….(i) \\ \text { and } & \frac{h c}{\lambda _2} =\phi+\frac{1}{2} m v^{2} \\ \text { or } & \frac{h c}{\lambda _2}-\phi =\frac{1}{2} m v^{2} \quad …….(ii) \end{aligned} $
Now, we divide Eq. (i) by (ii) Eq.
or
$ \frac{\frac{h c}{\lambda _1}-\phi}{\frac{h c}{\lambda _2}-\phi}=\frac{\frac{1}{2} m \times 4 v^{2}}{\frac{1}{2} m v^{2}}=4 $
or
$ \frac{h c}{\lambda _1}-\phi=\frac{4 h c}{\lambda _2}-4 \phi $
$\text { or } \phi=\frac{1}{3} h c (\frac{4}{\lambda _2}-\frac{1}{\lambda _1}) $
$=\frac{1}{3} \times 1240 (\frac{4 \times 350-540}{350 \times 540}) $
$\text { or } \phi =1.8 $ $eV$
According the assumption II, correct option is (c).
BETA
