JEE Mains · Chemistry · STD 11 - 2. structure of atom
When light of wavelength \(248\, nm\) falls on a metal of threshold energy \(3.0 \,eV\), the de-Broglie wavelength of emitted electrons is ............. \(\mathring A\) (Round off to the Nearest Integer). [Use : \(\sqrt{3}=1.73, h =6.63 \times 10^{-34} Js\) \(m _{ e }=9.1 \times 10^{-31} kg ; c =3.0 \times 10^{8} ms ^{-1}\) \(\left.1 eV =1.6 \times 10^{-19} J \right]\)
- A \(7\)
- B \(9\)
- C \(12\)
- D \(18\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
Energy incident \(=\frac{ hc }{\lambda}\) \(=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{248 \times 10^{-9} \times 1.6 \times 10^{-19}} eV\) \(=\frac{6.63 \times 3 \times 100}{248 \times 1.6}\) \(=0.05 eV \times 100=5 eV\) Now using \(E =\phi+ K . E\) \(5=3+ K . E\)…
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