In an experiment on photoelectric emission from a metallic surface ,wavelength of incident light `2xx10^(-7)` m and stopping potential is 2.5V .The threshold frequency of the metal (in Hz) approximately
(Charge of electrons `e=1.6xx10^(-19)`C, Planck's constant ,`h=6.6xx10^(-34)J-s)`

Options

A. `12xx10^(15)`
B. `9xx10^(15)`
C. `9xx10^(14)`
D. `12xx10^(13)`

Answer

C

Explanation

In a photoelectric emisision ,
Wavelength ,`lamda=2xx10^(-7)`
Stopping potential `V_(o)=2.5`
Consider the equation
,`eV_(0)=hv-hv_(0)`
From the above equation threshold frequency can be calculated a`v_(o)=v-(eVo)/(h)`s.
`impliesv_(o)=("C")/(lamda)-(eVo)/h(thereforec=vlamda)`
`(v_(o)=3xx10^(8))/(2xx10^(-7))-((1.6xx10^(-19))(2.5))/(6.6xx10)`
`=(15-6)xx10^(14)V_(o)=9xx10^(14)H`
`sqrtV=B=(Z-A)implies sqrt(v)/(Z-A)=B`
where , v- Frequency of `K_(o)` line Z- Atomic number of elements