Question

metal sheet has a negative charge of 11.2 nC(1nC=10^-9).how many photons of UV are required to completely discharge the sheet by photoelectron ejection?.what minimum amount of energy must be absorbed by the sheet to effect this discharge? The threshold frequency of the metal is 4.5×10^15Hz​

Answer 1

Total number of photon will be $n=7\times 10^7photon$

Minimum energy will be $29.7\times 10^{-19}j$

Explanation:

We have given that charge on sheet $Q=11.2nC=11.2\times 10^{-9}C$

Charge on photon $e=1.6\times 10^{-19}C$

We know that charge is given by

$Q=ne$

$11.2\times 10^{-9}=n\times 1.6\times 10^{-19}$

$n=7\times 10^7electron$

We have given threshold frequency

$f=4.5\times 10^{15}Hz$

Plank's constant $h=6.6\times 10^{-34}J-s$

So energy is given by

$E=h\nu =6.6\times 10^{-34}\times 4.5\times 10^{15}=29.7\times 10^{-19}j$

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Answer 2

Number of photons of UV required to completely discharge the sheet by photoelectron ejection=$7\times 10^{10}$

Minimum amount of energy absorbed by the sheet to effect this discharge =$2.98\times 10^{-18} J$

Explanation:

Charge on metal sheet=$11.2n C$

1nC=$10^{-9} C$

Therefore, charge on metal sheet=$11.2\times 10^{-9} C$

We know that

$q=ne$

Charge on photon=1e-=$1.6\times 10^{-19} C$

Substitute the values then we get

$11.2\times 10^{-9}=n(1.6\times 10^{-19})$

$n=\frac{11.2\times 10^{-9}}{1.6\times 10^{-19}}$

$n=7\times 10^{10}$

Hence, number of photons of UV required to completely discharge the sheet by photoelectron ejection=$7\times 10^{10}$

Threshold frequency of the metal=$4.5\times 10^{5}$ Hz

Minimum amount of energy absorbed by the sheet to effect this discharge is given by

$E_0=h\nu_0$

Where

$E_0$=Minimum amount of energy required by electron to eject from the metal surface

$\nu_0$=Threshold frequency of the metal

$h=6.63\times 10^{-34} Js=$Plank's constant

Substitute the values in the formula then we get

$E_0=6.63\times 10^{-34}\times 4.5\times 10^{15}=2.98\times 10^{-18} J$

Hence, minimum amount of energy absorbed by the sheet to effect this discharge=$2.98\times 10^{-18} J$

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