In a photoemissive cell with exciting wavelength λ , the fastest electron has speed v. If the exciting wavelength is changed to
$3lambda /4$
, the speed of the fastest emitted electron will be

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$v;{(3/4)^{1/2}}$

$v;{(4/3)^{1/2}}$

$Less,than,v;{(4/3)^{1/2}}$

Greater than $v;{(4/3)^{1/2}}$

answer is D.

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Detailed Solution

$hnu - {W_0} = frac{1}{2}mv_{max }^2 Rightarrow frac{{hc}}{lambda } - frac{{hc}}{{{lambda _0}}} = frac{1}{2}mv_{max }^2$
$Rightarrow hcleft( {frac{{{lambda _0} - lambda }}{{lambda {lambda _0}}}} right) = frac{1}{2}mv_{max }^2 Rightarrow {v_{max }} = sqrt {frac{{2hc}}{m}left( {frac{{{lambda _0} - lambda }}{{lambda {lambda _0}}}} right)}$
When wavelength is λ and velocity is v, then
$v = sqrt {frac{{2hc}}{m}left( {frac{{{lambda _0} - lambda }}{{lambda {lambda _0}}}} right)}$ ........(i)
When wavelength is $frac{{3lambda }}{4}$ and velocity is v’ then
$v' = sqrt {frac{{2hc}}{m}left[ {frac{{{lambda _0} - (3lambda /4)}}{{(3lambda /4) times {lambda _0}}}} right]}$ .......(ii)
Divide equation (ii) by (i), we get
$frac{{v'}}{v} = sqrt {frac{{[{lambda _0} - (3lambda /4)]}}{{frac{3}{4}lambda {lambda _0}}} times frac{{lambda {lambda _0}}}{{{lambda _0} - lambda }}}$
$v' = v{left( {frac{4}{3}} right)^{1/2}}sqrt {frac{{[{lambda _0} - (3lambda /4)]}}{{{lambda _0} - lambda }}} ,i.e.,v' > v{left( {frac{4}{3}} right)^{1/2}}$