Answer to Question #227995 in Molecular Physics | Thermodynamics for Ezekiel

Question

A laser with power output 2.00microwatts
And a wavelength of. 400nano meter
Is projected into a calcium metal
Find how many electron is released per second
Find power of electro. If binding energy is 2.71ev

Answer 1

A The energy for a single photon of 400 nm wavelength is:

$E=h\nu=\dfrac{hc}{\lambda}=$

$\dfrac{(6.63 ×10^{-34} m^2 kg s^{-1}) × (3.0 x 10^8 m s^{-1})}{400 × 10^{-9} m}=4.97 ×10^{-19} J$

Power output of 2.00 mW means there are 2.00 mJ of energy deposited on the calcium metal per second through absorption of 400 nm photons. 2.00 mJ of energy corresponds to $\dfrac{2×10^{-3}J}{4.97×10−19J}=4.02×10^{15 }$photons.

Each photon will eject one electron, therefore, $4.02×10^{15 }$ electrons will be ejected per second.

(b) The energy that is carried away by each electron is the kinetic energy that it has after being ejected:

$KE_{e}=h\nu-\Phi=(4.97×10^{-19} J )- (2.71 eV×\dfrac{1.60 × 10^{-19} J}{1 eV})=\\6.34 ×10^{-20} J$

The power P_e carried away by the electrons is the energy that is carried away by all the ejected electrons per second:

$P_{e}=(4.02×10^{15} s^{-1}) ×(6.34 × 10^{-20} J)=0.255 mW$