What amount of energy sould be added to an electron to reduce its de Broglie wavelength `lamda_1=550nm` incident on it, causing the ejection of photoelectrons for which the stopping potential is `V_(s1)=0.19V`. If the radiation of wavelength `lamda_2=190nm` is now incident on the surface, (a) calculate the stopping potential `V_(S2)`, (b) the work function of the surface, and (c) the threshold frequency for the surface.

What amount of energy sould be added to an electron to reduce its de B

1.	What amount of energy sould be added to an electron to reduce its de Broglie wavelength `lamda_1=550nm` incident on it, causing the ejection of photoelectrons for which the stopping potential is `V_(s1)=0.19V`. If the radiation of wavelength `lamda_2=190nm` is now incident on the surface, (a) calculate the stopping potential `V_(S2)`, (b) the work function of the surface, and (c) the threshold frequency for the surface.
Answer» We know that, de Broglie wavelength `lamda=(h)/(mv)` and `R=(1)/(2)mv^2` `lamda=(h)/(sqrt(2mE))` In first case, `100xx10^(-12)=(h)/(sqrt(2mE_(1))` …(i) In second case, `50xx10^(-12)=(h)/(sqrt(2mE_(2))` ..(ii) Dividing Eqs. (i) by (ii), we get `2=sqrt(((E_2)/(E_1)))` or `E_2=4E_1` So, energy to be added `=4E_1-E_1=3E_1` Now, `(h)/(sqrt(2mE_1))=100xx10^(-12)` or `sqrt(2mE_1)=(6.625xx10^(-34))/(10^(-10))` or `sqrt(2mE_1)=6.625xx10^(-24)` or `E_1=((6.625xx10^(-24))^(2))/(2xx(9.1xx10^(-31)))=150eV` There, energy added `=3E_1=450eV`

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