Stopping potentials of 24, 100, 110 and 115 | vedclass

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(C) The stopping potentials correspond to the kinetic energies of photoelectrons ejected from different shells. The maximum kinetic energy corresponds

Vedclass · Accepted Answer

$0.163 \ \mathring{A}$

Vedclass · Answer

(C) The stopping potentials correspond to the kinetic energies of photoelectrons ejected from different shells. The maximum kinetic energy corresponds to the least tightly bound electrons,while the minimum stopping potential $(24 \ kV)$ corresponds to the most tightly bound electrons,i.e.,the $K$-shell.The energy of the $K$-shell is $E_K = -24 \ keV$ (relative to the vacuum level,binding energy is $24 \ keV$).The energy of the $L$-shell is $E_L = -100 \ keV$ (binding energy is $100 \ keV$).The energy difference for the $K_{\alpha}$ transition (from $L$-shell to $K$-shell) is:$\Delta E = E_L - E_K = 100 \ keV - 24 \ keV = 76 \ keV = 76 	imes 10^3 \ eV$.The wavelength $\lambda$ of the emitted $X-ray$ photon is given by:$\lambda = \frac{hc}{\Delta E} \approx \frac{12400 \ eV \cdot \mathring{A}}{76 	imes 10^3 \ eV} \approx 0.163 \ \mathring{A}$.

Stopping potentials of $24, 100, 110$ and $115 \ kV$ are measured for photoelectrons emitted from a certain element (with $24 \ kV$ minimum) when it is irradiated with monochromatic $X-rays$. If this element is used as a target in an $X-ray$ tube,what will be the wavelength of the $K_{\alpha}$ line?

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