Question 219: Dual Nature of Matter and Radiation

Question

When photons of wavelength $${\lambda _1}$$ are incident on an isolated sphere, the corresponding stopping potential is found to be V. When photons of wavelength $${\lambda _2}$$ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength $${\lambda _3}$$ is used then find the stopping potential for this case :

Accepted Answer

We know,
Einstein's photoelectric equation,
$$eV = {{hc} \over \lambda } - {\phi _0}$$
and $${\phi _0}$$, $${{hc} \over {{\lambda _0}}}$$, where $${{\lambda _0}}$$ is the threashhold wavelength.
$$ 	herefore $$   In first case,
eV $$ = {{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _0}}}$$      . . .(1)
and in second case,
3 eV $$ = {{hc} \over {{\lambda _2}}} - {{hc} \over {{\lambda _0}}}$$       . . .(2)
Now, let slopping potential = V
1
when light of wavelength $$\lambda $$
3
is used then,
eV
1
$$ = {{hc} \over {{\lambda _3}}} - {{hc} \over {{\lambda _0}}}$$       . . .(3)
From (1) and (2) get,
$$3\left( {{{hc} \over {{\lambda _1}}} - {{hc} \over {{\lambda _0}}}} ight) = {{hc} \over {{\lambda _2}}} - {{hc} \over {{\lambda _0}}}$$
$$ \Rightarrow $$   $${{3hc} \over {{\lambda _1}}}$$ $$-$$ $${{hc} \over {{\lambda _2}}}$$ $$=$$ $${{2hc} \over {{\lambda _0}}}$$
$$ \Rightarrow $$   $${{hc} \over {{\lambda _0}}}$$ $$=$$ $${{3hc} \over {2{\lambda _1}}}$$ $$-$$ $${{hc} \over {2{\lambda _2}}}$$
Putting this value of $${{hc} \over {{\lambda _0}}}$$ in equation 3,
eV
1
$$ = {{hc} \over {{\lambda _3}}} - {{3hc} \over {2{\lambda _1}}} + {{hc} \over {2{\lambda _2}}}$$
$$ \Rightarrow $$   V
1
$$ = {{hc} \over e}$$ $$\left[ {{1 \over {{\lambda _3}}} - {3 \over {2{\lambda _1}}} + {1 \over {2{\lambda _2}}}} ight]$$

Answer

$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} - {1 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$$

Answer

$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$$

Answer

$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {2{\lambda _2}}} - {3 \over {2{\lambda _1}}}} \right]$$

Answer

$${{hc} \over e}\left[ {{1 \over {{\lambda _3}}} + {1 \over {2{\lambda _2}}} - {1 \over {{\lambda _1}}}} \right]$$

Dual Nature of Matter and Radiation

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