(1)Two light sources are used in a photoelectric experiment to determine the wor

Question

(1)Two light sources are used in a photoelectric experiment to determine the work function for a particular metal surface. When green light from a mercury lamp ( = 546.1 nm) is used, a stopping potential of 0.626 V reduces the photocurrent to zero.

(a) Based on this measurement, what is the work function for this metal? ______eV

(b) What stopping potential would be observed when using light from a red lamp ( = 626.0 nm)? ______V

(2) X-rays having an energy of 330 keV undergo Compton scattering from a target. The scattered rays are detected at 39.0° relative to the incident rays.

(a) Find the Compton shift at this angle.
_______m

(b) Find the energy of the scattered x-ray.
_____keV

(c) Find the energy of the recoiling electron.
_______keV

Explanation / Answer

part a:

as we know, only a light wave of energy greater than the work function of the metal will be able to generate photo currents

maximum energy of released photons=energy of the incident light wave-work function of the metal

in order to make the photo current zero, a opposite potential has to be proivded which will add to the work function of metal

and oppose the release of photons

for zero photo currents,

energy of incident light=work function + energy corresponding to stopping potential

==>h*c/wavelength=work function + e*potential applied

==>6.626*10^(-34)*3*10^8/(546.1*10^(-9))=work function + 1.6*10^(-19)*0.626

==>work function=3.64*10^(-19) -1.0016*10^(-19)=2.6384*10^(-19) J

in units of eV, work function=(2.6384*10^(-19))/(1.6*10^(-19))=1.649 eV

part b:

energy associated with stopping potential while using right of red lamp

=(h*c/wavelength)-work function

=(6.626*10^(-34)*3*10^8/(626*10^(-9)))-2.6384*10^(-19)=5.37*10^(-20) J

then stopping potential=5.37*10^(-20)/(1.6*10^(-19))=0.33562 volts

Q2.

Q2.

energy of incident ray=330 keV=330*10^3*1.6*10^(-19) J=5.28*10^(-14) J

then its wavelength=h*c/energy=6.626*10^(-34)*3*10^8/(5.28*10^(-14))=3.7648*10^(-12) m

then compton scattering=(h/(m*c))*(1-cos(theta))

where m=mass of electron=9.1*10^(-31) kg

hence compoton shift=(6.626*10^(-34)/(9.1*10^(-31)*3*10^8))*(1-cos(39))=5.4089*10^(-13) m

part b:

then wavelength of scattered x ray=original wavelength+compton shift

=3.7648*10^(-12) +5.4089*10^(-13)

=4.3057*10^(-12) m

then energy of scattered x ray=h*c/wavelength=(6.626*10^(-34)*3*10^8)/(4.3057*10^(-12))=4.6167*10^(-14) J=288.54 keV

part c:

energy of recoiled electron=eenrgy of incident x ray-energy of scattered x ray=330-288.54=41.458 keV

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.