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Question

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda} = \frac{12400 \; eV \cdot \mathring A}{4500 \; \mathring A} \approx 2.76 \;

Vedclass · Accepted Answer

$1.36$

Vedclass · Answer

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda} = \frac{12400 \; eV \cdot \mathring A}{4500 \; \mathring A} \approx 2.76 \; eV$.The radius of the circular path of a charged particle in a magnetic field is $R = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.Rearranging for $K$: $K = \frac{(qBR)^2}{2m}$.Given $q = 1.6 	imes 10^{-19} \; C$,$B = 2 	imes 10^{-3} \; T$,$R = 2 	imes 10^{-3} \; m$,and $m = 9.1 	imes 10^{-31} \; kg$:$K = \frac{(1.6 	imes 10^{-19} 	imes 2 	imes 10^{-3} 	imes 2 	imes 10^{-3})^2}{2 	imes 9.1 	imes 10^{-31}} = \frac{(6.4 	imes 10^{-25})^2}{18.2 	imes 10^{-31}} = \frac{40.96 	imes 10^{-50}}{18.2 	imes 10^{-31}} \approx 2.25 	imes 10^{-19} \; J$.Converting to $eV$: $K = \frac{2.25 	imes 10^{-19}}{1.6 	imes 10^{-19}} \approx 1.40 \; eV$.Using Einstein's photoelectric equation: $\phi = E - K = 2.76 \; eV - 1.40 \; eV = 1.36 \; eV$.

List-$I$	List-$II$
$(P)$ $2000 \, K$	$(1)$ The radiation at peak wavelength can lead to emission of photoelectrons from a metal of work function $4 \, eV$
$(Q)$ $3000 \, K$	$(2)$ The radiation at peak wavelength is visible to human eye.
$(R)$ $5000 \, K$	$(3)$ The radiation at peak emission wavelength will result in the widest central maximum of a single slit diffraction.
$(S)$ $10000 \, K$	$(4)$ The power emitted per unit area is $1/16$ of that emitted by a blackbody at temperature $6000 \, K$.
	$(5)$ The radiation at peak emission wavelength can be used to image human bones.

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