Question
Download Solution PDFIf the kinetic energy of an electron released in a photoelectric emission process is quadrupled then wavelength of the electron become-
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCONCEPT:
The wavelength of an electron due to its motion is called the de-Broglie wavelength of the electron.
The de-Broglie wavelength of the electron (λe) is given by:
\({\lambda _e} = \frac{h}{{\sqrt {2m\;\left( {KE} \right)} }}\)
Where m is the mass of the electron, h is Planck constant and KE is the kinetic energy of the electron
EXPLANATION:
Given that: KE is 4 times the initial KE
KE’ = 4 KE
\({\lambda _e} = \frac{h}{{\sqrt {2m\;\left( {KE} \right)} }}\)
the final wavelength of an electron
\(\left( {{{\lambda '}_e}} \right) = \;\frac{h}{{\sqrt {2m\;\left( {4\;KE} \right)} }} = \frac{1}{2} \times \;\frac{h}{{\sqrt {2m\;\left( {KE} \right)} }} = \frac{{{\lambda _e}}}{2}\)
Thus the wavelength of the electron becomes half of the initial value. So option 2 is correct.
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