ફોટો ઈલેક્ટ્રિક અસર સપાટી પર $5.5 \times 10^8\ MHz$ અને $4.5 \times 10^8\ MHz$ માટે મળે છે. જો ઉત્સર્જિત ફોટો ઈલેક્ટ્રોનની મહત્તમ ગતિઊર્જાનો ગુણોત્તર $1:5$ હોય તો ધાતુની સપાટી માટે થ્રેશોલ્ડ આવૃત્તિ .....

Question

A$7.55 \times 10^8\ MHz$
B$4.57 \times 10^8\ MHz$
C$9.35 \times 10^8\ MHz$
D$5.75 \times 10^8\ MHz$

Diffcult

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Solution

d
$\,E\,\, = \,\,\frac{1}{2}m\upsilon _{\max }^2\,\, = \,\,h{f}\,\, - \,\,h{{f}_0}\,$ $ = \,\,h\,({f}\,\, - \,\,{{f}_0})\,\,\,\,{E_1}\, = \,\,h\,({{f}_1}\, - \,\,{{f}_0})$ અને

${E_2}\, = \,\,h({{f}_2}\, - \,\,{{f}_0})$

${{f}_1}\, = \,\,5.5\,\, \times \,\,{10^8}\,MHz\,\, = \,\,5.5\,\, \times \,\,{10^{14}}\,Hz,$

${{f}_2}\, = \,\,4.5\,\, \times \,\,{10^8}\,MHz\,\, = \,\,4.5\,\, \times \,\,{10^{14}}\,MHz$

$\frac{{{E_1}}}{{{E_2}}}\,\, = \,\,\,\frac{{{f_1}\, - \,\,{f_0}}}{{{f_2}\, - \,\,{f_0}}}\,\,$ પરંતુ

$\frac{{{{\rm{E}}_{\rm{1}}}}}{{{{\rm{E}}_{\rm{2}}}}}\,\, = \,\,\frac{1}{5}$ હોવાથી

$\therefore \,\,\,\frac{{\rm{1}}}{{\rm{5}}}\,\, = \,\,\,\frac{{5.5\,\, \times \,\,{{10}^{14}}\, - \,\,{{f}_0}}}{{4.5\,\, \times \,\,{{10}^{14}}\, - \,\,{{f}_0}}}\,\,\,\,\,$

$\therefore \,\,\,4.5\,\, \times \,\,{10^{14}}\, - \,\,{{f}_0}\,\, = \,\,27.5\,\, \times \,\,{10^{14}}\, - \,\,5\,{{f}_0}$

$\therefore \,\,4{{f}_0}\, = \,\,23.0\,\, \times \,\,{10^{14}}$

$\therefore \,\,{{f}_0}\, = \,\,\,\frac{{23\,\, \times \,\,{{10}^{14}}}}{4}\,\, = \,\,5.75\,\, \times \,\,{10^{14}}\,Hz\,$

$ = \,\,5.75\,\,\, \times \,\,{10^8}\,Hz$

લીસ્ટ $I$	લીસ્ટ $II$
$A$ પલાન્ક અચળાંક $( h )$	$I$ $\left[ M ^1 L ^2 T ^{-2}\right]$
$B$ સ્ટોપિંગ પોટેન્શિયલ $( Vs )$	$II$ $\left[ M ^1 L ^1 T ^{-1}\right]$
$C$ કાર્ય વિધેય $(\phi)$	$III$ $\left[ M ^1 L ^2 T ^{-1}\right]$
$D$ વેગમાન $( p )$	$IV$ $\left[ M ^1 L ^2 T ^{-3} A ^{-1}\right]$

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