When radiation of the wavelength `lamda`is incident on a metallic surface ,the stopping potential is 4.8 V. If the same surface is illuminated with radiation of double the wavelength ,then the stopping potential becomes 1.6 V.Then the threshold wavelength for the surface is

Options

A. `2lamda`
B. `4lamda`
C. `6lamda`
D. `8lamda`

Answer

B

Explanation

First stopping potenital ,`V_(o)=4.8V.`
Second stopping potential ,`V_(o)`=1.6V
The expression for stopping potential is given as ,
`V_(o)=(Hc)/(e)((1)/(lamda)-(1)/(lamda))`
Where
`lamda_(o)`Threshold wavelength
The threshod wavelength can be calculated as ,
`(4.8)/(1.6)=((hc)/(e)((1)/(lamda)-(1)/(lamda_(o))))/((hc)/(e)((1)/(2lamda)-(1)/(lamda_(0))))`
`implies3=(((1)/(2lamda)-(1)/lamda_(0)))/(((1)/(2lamda)-(1)/(lamda_(o)))`
`implies (3)/(2lamda)-(3)/(lamda_(o))=(1)/(lamda)-(1)/(lamda_(0)`
`implies (1)/(lamda_(o))-(3)/(lamda_(0))=(1)/(lamda)-(3)/(2lamda)`
`implies -(2)/(lamda _(o))=(2-3)/(2lamda)`
`implies (-2)/lamda_(o)=(-1)/(2lamda)`
`thereforelamda_(o)=4lamda`