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In the arrangement shown in figure y=1.0...

In the arrangement shown in figure y=1.0 nm, d=0.24 mm and D=1.2m. The work function of the material of the emitter is 2.2 eV. If stopping potential is 0.3 x, then value of x (in V) is

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If `lamda` is the wavelength emitted by the source S, then fringe width
`beta=1y=(D lamda)/d`
`:.lamda=(2yd)/D`
`=(2xx(1xx10^(-3))xx(0.24xx10^(-3)))/1.2`
`implies4xx10^(-6)m`
By Einstein equation
`hf=W_(0)+eV`
or `(hc)/(lamda)=W_(0)+eV`
`:.V=(hc)/(elamda)-(W_(0))/e`
`=((6.63xx10)^(-34)xx(3xx10^(8)))/((1.6xx10^(-19))xx(0.4xx10^(-6)))-(2.2xx1.6xx10^(-19))/(1.6xx10^(-19))`
`=0.9V`
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