Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
+1 vote

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.

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

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App