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Question Number 55748 by necx1 last updated on 03/Mar/19

Commented by necx1 last updated on 03/Mar/19

$$pleasehelp$$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Mar/19

$$h\nu =h{\nu }_o+\frac{1}{2}{mv}_{max}^2$$$$\frac{hc}{\lambda }=\frac{hc}{{\lambda }_o}+\frac{1}{2}{mv}_{max}^2$$$$\lambda =2000\times {10}^{-10}meter$$$$ithink{\lambda }_0=4400\times {10}^{-10}meter$$$${(K.E)}_{max}=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$$$$b)★plscheck...$$$$changeofkineticenergy=workdone$$$$charge\times potentialdifference=(\frac{1}{2}{mc}^2-0)$$$$ev=\frac{1}{2}{mc}^2$$$$v=\frac{{mc}^2}{2e}★$$$$$$

Commented by necx1 last updated on 03/Mar/19

$$exaxtly!Thankyousomuch$$

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