Saved Bookmarks
| 1. |
When radiation of wavelength `lambda ` is incident on a metallic surface , the stopping potential is `4.8 "volts"` . If the same surface is illuminated with radiation of double the wavelength , then the stopping potential becomes `1.6 "volts"`. Then the threshold wavelength for the surface isA. 2`lambda`B. 4`lambda`C. 6`lambda`D. 8`lambda` |
|
Answer» Correct Answer - B `(hc)/e (1/lambda-1/lambda_0) = V_0` `rArr "hc"/e (1/lambda - 1/lambda_0) =4.8` …(i) and `(hc)/e (1/(2lambda)-1/lambda_0) =1.6` …(ii) From equation (i) by (ii) , `((1/lambda-1/lambda_0))/((1/(2lambda)-1/(lambda_0)))=4.8/1.6 rArr lambda_0 = 4lambda` |
|