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Light of wavelength $0.6\: \mu m$ from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is $0.5\: V$ and with light of wavelength $0.4\: \mu m$ from a sodium lamp, the stopping potential is $1.5\: V$. With this data, the value of $ \large\frac{h}{e}$ will be

$\begin {array} {1 1} (a)\;4 \times 10^{-19}\: Vs & \quad (b)\;0.25 \times 10^{-15}\: Vs \\ (c)\;4 \times 10^{-15}\: Vs & \quad (d)\;4 \times 10^{-8}\: Vs \end {array}$

 

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