The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function \phi, when light frequency v incidents on it is \lambda. Then v= ( h - Planck's constant, m - mass of electron)

\frac{2 \phi}{h}-\frac{h}{m \lambda^2}

\frac{2 \phi}{ {\h}}+ { {\h}}/{ {\m} \lambda^2}

\frac{\phi}{ {\h}}+ { {\h}}/{2 {\m} \lambda^2}

\frac{\phi}{ {\h}}- { {\h}}/{2 {\m} \lambda^2}

AP EAPCET 15th May 2023 Shift 2 Solved Paper

Section: Physics

Question : 115 of 160

Marks: +1, -0

The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function φ, when light frequency v incidents on it is
λ. Then v=
( h - Planck's constant, m - mass of electron)

‌
2φ
h
−‌
h
mλ2
‌
2φ
h
+‌
h
mλ2
‌
φ
h
+‌
h
2mλ2
‌
φ
h
−‌
h
2mλ2

Validate

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