When photons of energy 4.25 eV strike the surface of a metal "A", the ejected photoelectrons have ma

Question

When photons of energy 4.25 eV strike the surface of a metal "A", the ejected photoelectrons have maximum kinetic energy, TA expressed in eV and de Broglie wavelength &#955;. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB = (TA - 1.50 eV). If the de Broglie wavelength of these photoelectrons is &#955;B = 2&#955;A, then :

Accepted Answer

4.25 = TA + WA4.70 = TB + WBTB = TA - 1.5&#955; = & &#955;A = &#955;B &#8658; TA = 4TB

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