logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

The threshold frequency for a metallic surface corresponds to energy of $6.2\: eV$ and the stopping potential for a radiation incident on this surface is $5V$. The incident radiation lies in

$\begin {array} {1 1} (a)\;Ultraviolet \: region & \quad (b)\;Infrared\: region \\ (c)\;Visible\: region & \quad (d)\;X-ray\: region \end {array}$

 

Can you answer this question?
 
 

1 Answer

0 votes
Ans : (a) Ultraviolet region
$hv_{\circ} = 6.2\: eV \:\: \: \: \: \: \: \: \: eV_{\circ} = 5eV$
So,$ hv = hv_{\circ} + eV_{\circ} = 6.2 + 5 = 11.2\: eV$
$ \Rightarrow hc/\lambda = 11.2$
$ \Rightarrow \lambda = hc/11.2 = 1108.9 A^{\circ}$ which is in ultraviolet region.

 

answered Dec 23, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

Related questions

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