Question

If the critical wavelength of tungsten is 26008, what is the energy of a quantum at this wavelength in eV? Further, calculate the wavelength necessary to produce photo electrons from tungsten having twice the kinetic energy of those produced at 22004.

JEE/Engineering Exams

Chemistry

Solution

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Witical wareleng/n =w wavelergan when k.E is novmal=w, when ( k: E ) is doubled ( =w_{2} ) ( k in_{1}=frac{1}{w_{1}}-frac{1}{w_{0}}, k in_{2}=2 k E_{1} ) ( k in_{2}=frac{1}{w_{2}}-frac{1}{w}=frac{2}{w_{1}}-frac{2}{w^{0}} ) ( Rightarrow frac{1}{w_{2}}=frac{2}{w_{1}}-frac{1}{w_{0}} ) ( frac{1}{w^{2}}=frac{2}{2200}-frac{1}{2600}=frac{2600-1100}{2860000} ) ( begin{aligned} frac{1}{w^{2}}=frac{1500}{2860000} Rightarrow w_{2}=& frac{2860000}{1500} w_{2}=& 1906.667 text { A } end{aligned} )