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When a certain metal was irradiated with light of frequency $1.6\times 10^{16}Hz$,the photoelectrons emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light of frequency $1.0\times 10^{16}$Hz.Calculate $V_0$(threshold frequency) for the metal.

$\begin{array}{1 1}(a)\;3\times 10^{15}Hz&(b)\;4\times 10^{10}Hz\\(c)\;4\times 10^{15}Hz&(d)\;2\times 10^{15}Hz\end{array}$

1 Answer

$KE_1=h(V_1-V_0)$------(i)
$KE_2=h(V_2-V_0)=\large\frac{KE_1}{2}$------(i)
Dividing equations (ii) by (i) we have
$\therefore \large\frac{V_2-V_0}{V_1-V_0}=\frac{1}{2}$
$\large\frac{1.0\times 10^{16}-V_0}{1.6\times 10^{16}-V_0}=\frac{1}{2}$
$2.0\times 10^{16}-2V_0=1.6\times 10^{16}-V_0$
$V_0=4\times 10^{15}Hz$
Hence (c) is the correct answer.
answered Jan 16, 2014 by sreemathi.v
 

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