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Structure of Atom question

2021 · 16 Mar · Shift 1 · Q21
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Structure of Atom question

2021 · 16 Mar · Shift 1 · Q21

JEE MainChemistryStructure of AtomNumerical+4 / −1
When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is ‾\underline{\hspace{2cm}}​Ao\mathop A\limits^oAo​. (Round off to the Nearest Integer). [ Use : 3\sqrt 33​= 1.73, h = 6.63 ×\times× 10 −-− 34 Js me = 9.1 ×\times× 10 −-− 31 kg; c = 3.0 ×\times× 108 ms −-− 1; 1eV = 1.6 ×\times× 10 −-− 19 J]
Numerical answer
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Correct answer: 9

  1. Use Einstein’s photoelectric equation

For emitted electrons, K.E.=hν−ϕ=hcλ−ϕK.E. = h\nu - \phi = \frac{hc}{\lambda} - \phiK.E.=hν−ϕ=λhc​−ϕ where threshold energy ϕ=3.0 eV\phi = 3.0\,\text{eV}ϕ=3.0eV and incident wavelength λ=248 nm=248×10−9 m\lambda = 248\,\text{nm} = 248\times 10^{-9}\,\text{m}λ=248nm=248×10−9m

  1. Calculate energy of incident photon

E=hcλE = \frac{hc}{\lambda}E=λhc​ Substitute the given values: E=(6.63×10−34)(3.0×108)248×10−9E = \frac{(6.63\times 10^{-34})(3.0\times 10^8)}{248\times 10^{-9}}E=248×10−9(6.63×10−34)(3.0×108)​

E=19.89×10−26248×10−9E = \frac{19.89\times 10^{-26}}{248\times 10^{-9}}E=248×10−919.89×10−26​

E≈8.02×10−19 JE \approx 8.02\times 10^{-19}\,\text{J}E≈8.02×10−19J

Convert into eV: E=8.02×10−191.6×10−19≈5.01 eVE = \frac{8.02\times 10^{-19}}{1.6\times 10^{-19}} \approx 5.01\,\text{eV}E=1.6×10−198.02×10−19​≈5.01eV

  1. Find maximum kinetic energy of emitted electron

K.E.=5.01−3.0=2.01 eVK.E. = 5.01 - 3.0 = 2.01\,\text{eV}K.E.=5.01−3.0=2.01eV

In joules, K.E.=2.01×1.6×10−19K.E. = 2.01\times 1.6\times 10^{-19}K.E.=2.01×1.6×10−19 K.E.≈3.216×10−19 JK.E. \approx 3.216\times 10^{-19}\,\text{J}K.E.≈3.216×10−19J

  1. Relate kinetic energy to momentum

For electron, K.E.=p22mK.E. = \frac{p^2}{2m}K.E.=2mp2​ So, p=2m K.E.p = \sqrt{2m\,K.E.}p=2mK.E.​

Using de-Broglie relation, λdB=hp=h2m K.E.\lambda_{dB} = \frac{h}{p} = \frac{h}{\sqrt{2m\,K.E.}}λdB​=ph​=2mK.E.​h​

  1. Substitute values

λdB=6.63×10−342×9.1×10−31×3.216×10−19\lambda_{dB} = \frac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 3.216\times 10^{-19}}}λdB​=2×9.1×10−31×3.216×10−19​6.63×10−34​

First, denominator inside root: 2×9.1×3.216=58.53122\times 9.1\times 3.216 = 58.53122×9.1×3.216=58.5312 So, 2mK.E.=58.5312×10−50=5.85312×10−492mK.E. = 58.5312\times 10^{-50} = 5.85312\times 10^{-49}2mK.E.=58.5312×10−50=5.85312×10−49

Now, 5.85312×10−49=5.85312×10−24.5\sqrt{5.85312\times 10^{-49}} = \sqrt{5.85312}\times 10^{-24.5}5.85312×10−49​=5.85312​×10−24.5

5.85312≈2.42\sqrt{5.85312} \approx 2.425.85312​≈2.42 Thus, p≈2.42×10−24.5p \approx 2.42\times 10^{-24.5}p≈2.42×10−24.5

Since 10−24.5=3.16×10−2510^{-24.5} = 3.16\times 10^{-25}10−24.5=3.16×10−25, p≈2.42×3.16×10−25≈7.65×10−25 kg m s−1p \approx 2.42\times 3.16\times 10^{-25} \approx 7.65\times 10^{-25}\,\text{kg m s}^{-1}p≈2.42×3.16×10−25≈7.65×10−25kg m s−1

Hence, λdB=6.63×10−347.65×10−25≈0.867×10−9 m\lambda_{dB} = \frac{6.63\times 10^{-34}}{7.65\times 10^{-25}} \approx 0.867\times 10^{-9}\,\text{m}λdB​=7.65×10−256.63×10−34​≈0.867×10−9m

λdB≈8.67×10−10 m\lambda_{dB} \approx 8.67\times 10^{-10}\,\text{m}λdB​≈8.67×10−10m

  1. Convert into angstrom

Since 1 A˚=10−10 m1\,\text{\AA} = 10^{-10}\,\text{m}1A˚=10−10m we get λdB≈8.67 A˚\lambda_{dB} \approx 8.67\,\text{\AA}λdB​≈8.67A˚

Rounded to nearest integer: 9\boxed{9}9​

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