When light of wavelength 248nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is …….. A∘. (Round off to the Nearest Integer). [Use: 3=1.73,h=6.63×10−34 ] Js[ me=9.1×10−31kg;c=3.0×108ms−1;1eV=1.6×10−19J ]

Energy of incident radiation =hcλ6.63×10−34×3×108248×10−90.0802×10−178.02×10-19J(w)work function = 3ev = 3×1.602×10−19J =4.806×10-19J K.E of photo e- = hv-w =8.02×10-19- 4.806×10−19 =3.214×10−19J debrogli wavelength λ= h2mKE λ=6.63× 10−342×9.1×10−31×3.214×10−19 λ=6.63× 10−3458.49×10−50=6.637.64×10−9 =0.86×10−9m λ=8.6A0

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When light of wavelength 248nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is …….. $A^{\circ} .$ (Round off to the Nearest Integer). [Use: $\sqrt{3} = 1.73, h = 6.63 \times 10^{- 34}$ ] Js
[ $m_{e} = 9.1 \times 10^{- 31} k g; c = 3.0 \times 10^{8} m s^{- 1}; 1 e V = 1.6 \times 10^{- 19} J$ ]

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Detailed Solution

Energy of incident radiation $= \frac{hc}{λ}$

$\begin{array}{l} \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{248 \times 10^{- 9}} \\ 0.0802 \times 10^{- 17} \\ 8.0 {2×10}^{-19} J \\ (w) work function = 3ev = 3 \times 1.602 \times 10^{- 19} J \\ = {4.806×10}^{-19} J \\ {K.E of photo e}^{-} = hv-w \\ {=8.02×10}^{-19} - 4 .806 \times 10^{- 19} \\ = 3.214 \times 10^{- 19} J \\ debrogli wavelength λ= \frac{h}{\sqrt{2mKE}} \\ λ= \frac{6.63 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 3.214 \times 10^{- 19}}} \\ λ= \frac{6.63 \times 10^{- 34}}{\sqrt{58.49 \times 10^{- 50}}} = \frac{6.63}{7.64} \times 10^{- 9} \\ = 0.86 \times 10^{- 9} m \\ {λ=8.6A}^{0} \end{array}$