The vapours of Hg absord some electron accelerated by a potiential diff. of 4.5 volt as a result of which light is emitted . If the full energy of single incident `e^(-)` is supposed to be converted into light emitted by single Hg atom , find the wave no. of the light

Answer

`3.63xx10^(6)m^(-1)`

Explanation

`4.5 eV = (1240)/(lambda(nm))" "(1)/(lambda)=(4.5)/(1240)`
`(1)/(lambda)=0.0036nm^(-1)" "1nmrarr10^(-9)m^(-1)`
`0.0036nm^(-1)" "1nmrarr10^(-9)m^(-1)`