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  • The     of the photoelectron is E when the incident wavelength is <img alt="\lambda" src="http

The  K\cdot E   of the photoelectron is E when the incident wavelength is \lambda  . To increse the  K\cdot E of the electron to 2E, the incident wavelength must be

Option: 1

2\lambda

 


Option: 2

\frac{\lambda }{2}


Option: 3

\frac{hc \lambda }{E\lambda +hc}


Option: 4

\frac{hc \lambda }{Ec+h\lambda }


Answers (1)

best_answer

E=\frac{hc}{\lambda }-\phi _{0}

similarly \Rightarrow 2E=\frac{hc}{{\lambda }'}-\phi _{0}

solving we get {{\lambda }'}=\frac{hc\lambda }{E\lambda+hc}

Posted by

HARSH KANKARIA

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