The energy of photoelectrons emtted from a photo-snsitive plate is 1.56 eV if threshold waveelngth is `2,500 @A`,calculate the waveelngth of incident light.Given ,1ev`=1.6 xx 10^(-12) erg` and `h=6.62 xx 10^(-27)` erg s.

Explanation

K.E. of Photoelectrons = `K.E= 1.56 eV`
K.E=`1.56 xx 1.6xx 10^(-19)J`
Threshold Wavelength `lamda_0= 2500 Å`
`lamda=2500x×10^(-10)m`
Wavelength of incident light = `lamda = ?`
`K.E=hc(1/(lamda)-1/(lamda_0))`
`(1/(lamda)-1/(lamda_0))=(K.E.)/(hc)`
= `(1.56xx1.6xx10^(-19))/(6625xx10-^(34)xx3xx10^8)`
`(2.496)/( 19.875)xx10^(-19+34-8)`
`1/(lamda)-1/lamda=0.126xx10^7`
`1/(lamda)-1/(lamda_0)=0.126xx10^7`
`1/(lamda)=0.126xx10^7+1/(lamda_0)`
`1/(lamda) =0.126xx10^(7)+1/(25xx10^(-8))`
`=0.126xx10^7+0.04xx10^8`
`=0.126xx10^7+0.4xx10^7 `
`1/(lamda)=0.526xx10^7`
`lamda=1/(0.526xx10^7)`
`=1.901xx10^(-7)m=`
`1901xx10^(-10)m=1901Å`