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If the speed of photoelectrons is $10^4\: m/s$, what should be the frequency of the incident radiation on a potassium metal? (work function of potassium = $2.3 eV)$

$\begin {array} {1 1} (a)\;5.6 \times 10^{15}Hz & \quad (b)\;0.56 \times 10^{15}Hz \\ (c)\;0.56 \times 10^{17}Hz & \quad (d)\;5.6 \times 10^{17}Hz \end {array}$

 

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Ans : (b)
$E = W_o+ E_k$ where, $W_o$ is the work function and $E_k$ is the KE of the
liberated photoelectron
$W_o = 2.3\: eV = 3.68 \times 10^{-19} J$
$E_k = \large\frac{1}{2} mv^2 $
$= \large\frac{1}{2} \times 9 \times 10^{-31} \times (10^4 )^2J = 4.5 \times 10^{-23} J$
So, $E = (3.68 \times 10^{-19} + 4.5 \times 10^{-23} ) J$
$= 3.68045 \times 10^{-19}J$
Frequency, $v= \large\frac{E}{h} = \large\frac{3.68045 \times 10^{-19} }{ (6.63 \times 10^{-34})}$
$= 0.56 \times 10^{15}Hz$

 

answered Dec 31, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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