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Structure of Atom question

2021 · 20 Jul · Shift 2 · Q18
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Structure of Atom question

2021 · 20 Jul · Shift 2 · Q18

JEE MainChemistryStructure of AtomNumerical+4 / −1
The wavelength of electrons accelerated from rest through a potential difference of 40 kV is x ×\times× 10 −-− 12 m. The value of x is ‾\underline{\hspace{2cm}}​. (Nearest integer) Give : Mass of electron = 9.1 ×\times× 10 −-− 31 kg Charge on an electron = 1.6 ×\times× 10 −-− 19 C Planck's constant = 6.63 ×\times× 10 −-− 34 Js
Numerical answer
View written solutionFree

Correct answer: 6

  1. Use de Broglie wavelength relation

For an electron accelerated from rest through potential difference VVV,

λ=hp\lambda = \frac{h}{p}λ=ph​

and the gained kinetic energy is

eV=p22meV = \frac{p^2}{2m}eV=2mp2​

So,

p=2meVp = \sqrt{2meV}p=2meV​

Hence,

λ=h2meV\lambda = \frac{h}{\sqrt{2meV}}λ=2meV​h​

  1. Substitute given values

Given:

h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J s}h=6.63×10−34J s m=9.1×10−31 kgm = 9.1 \times 10^{-31}\,\text{kg}m=9.1×10−31kg e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}e=1.6×10−19C V=40 kV=4.0×104 VV = 40\,\text{kV} = 4.0 \times 10^4\,\text{V}V=40kV=4.0×104V

Therefore,

λ=6.63×10−342×9.1×10−31×1.6×10−19×4.0×104\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 4.0 \times 10^4}}λ=2×9.1×10−31×1.6×10−19×4.0×104​6.63×10−34​

  1. Calculate the denominator inside the square root

First the numerical part:

2×9.1×1.6×4.0=116.482 \times 9.1 \times 1.6 \times 4.0 = 116.482×9.1×1.6×4.0=116.48

Powers of 10:

10−31×10−19×104=10−4610^{-31} \times 10^{-19} \times 10^4 = 10^{-46}10−31×10−19×104=10−46

So,

2meV=116.48×10−46=1.1648×10−442meV = 116.48 \times 10^{-46} = 1.1648 \times 10^{-44}2meV=116.48×10−46=1.1648×10−44

Now take square root:

1.1648×10−44=1.1648×10−22\sqrt{1.1648 \times 10^{-44}} = \sqrt{1.1648} \times 10^{-22}1.1648×10−44​=1.1648​×10−22

1.1648≈1.079\sqrt{1.1648} \approx 1.0791.1648​≈1.079

Thus,

2meV≈1.079×10−22\sqrt{2meV} \approx 1.079 \times 10^{-22}2meV​≈1.079×10−22

  1. Now compute wavelength

λ=6.63×10−341.079×10−22\lambda = \frac{6.63 \times 10^{-34}}{1.079 \times 10^{-22}}λ=1.079×10−226.63×10−34​

λ≈6.15×10−12 m\lambda \approx 6.15 \times 10^{-12}\,\text{m}λ≈6.15×10−12m

So,

λ=x×10−12 m\lambda = x \times 10^{-12}\,\text{m}λ=x×10−12m

Hence,

x≈6.15x \approx 6.15x≈6.15

Nearest integer:

x=6\boxed{x = 6}x=6​

  1. Comparison with stored answer

Derived answer = 666

Stored correct answer = 666

So they agree.

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