Question

Hola ❤

⏺ A photo electric surface has work function 2eV, what is the velocity of the photoelectrons ejected by light of wavelength 3000Å.

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Answer 1

work function of electron means energy required to break the bond ,with which electron is bounded....

now here

so,here

total energy of light=hv=hc/y. (y refers to wavelength)=6.626×10^-34×3×10^8/3×10^3=6.626×10^(-34+8-3)

now,this energy is transferred to electron,which is

used in breaking the bond of electron ,and to give it KE

so

6.626×10^(-29)=2eV+1/2 mv^2

6.626×10^-29=2×1.6×10^-19+1/2 ×9.1×10^-31×v^2(converting eletron volt to joule,we multiply it by 1.6×10^-19)

6.626×10^-29=(10^-19)(3.2+9.1/2×10^-12×v^2)

6.626×10^-10=3.2+(9.1×10^-12×v^2)/2

2×{(6.626×10^-10)-3.2}=9.1××10^-12×v^2

V^2=(13.24×10^-10-6.4)/(9.1×10^-12)

V^2=6.4/9.1×10^-12

V^2=7×10^12(approx value)

V=2.6×10^6

Answer 2

Answer:

total energy of light=hv=hc/y. (y refers to wavelength)=6.626×10^-34×3×10^8/3×10^3=6.626×10^(-34+8-3)

now,this energy is transferred to electron,which is

used in breaking the bond of electron ,and to give it KE

so

6.626×10^(-29)=2eV+1/2 mv^2

6.626×10^-29=2×1.6×10^-19+1/2 ×9.1×10^-31×v^2(converting eletron volt to joule,we multiply it by 1.6×10^-19)

6.626×10^-29=(10^-19)(3.2+9.1/2×10^-12×v^2)

6.626×10^-10=3.2+(9.1×10^-12×v^2)/2

2×{(6.626×10^-10)-3.2}=9.1××10^-12×v^2

V^2=(13.24×10^-10-6.4)/(9.1×10^-12)

V^2=6.4/9.1×10^-12

V^2=7×10^12(approx value)

V=2.6×10^6