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Structure of Atom question

2013 · Shift 0 · Q23
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Structure of Atom question

2013 · Shift 0 · Q23

JEE MainChemistryStructure of AtomMCQ+4 / −1
Energy of an electron is given by E=−2.178×10−18J(Z2n2)E = - 2.178 \times {10^{ - 18}}J\left( {{{{Z^2}} \over {{n^2}}}} \right)E=−2.178×10−18J(n2Z2​). Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be (h = 6.62 × 10−34 Js and c = 3.0 × 108 ms−1)
  1. A
    2.816 × 10−7 m
  2. B
    6.500 × 10−7 m
  3. C
    8.500 × 10−7 m
  4. D
    1.214 × 10−7 m
View written solutionFree

Correct answer: D

  1. Given energy formula

For hydrogen-like atoms,

En=−2.178×10−18(Z2n2) JE_n = -2.178 \times 10^{-18} \left(\frac{Z^2}{n^2}\right) \text{ J}En​=−2.178×10−18(n2Z2​) J

For a hydrogen atom, Z=1Z=1Z=1.

So,

En=−2.178×10−181n2 JE_n = -2.178 \times 10^{-18} \frac{1}{n^2} \text{ J}En​=−2.178×10−18n21​ J
  1. Find energies of the two levels
  • For n=1n=1n=1:
E1=−2.178×10−18 JE_1 = -2.178 \times 10^{-18} \text{ J}E1​=−2.178×10−18 J
  • For n=2n=2n=2:
E2=−2.178×10−18×14E_2 = -2.178 \times 10^{-18} \times \frac{1}{4}E2​=−2.178×10−18×41​ E2=−0.5445×10−18 JE_2 = -0.5445 \times 10^{-18} \text{ J}E2​=−0.5445×10−18 J
  1. Energy required for excitation from n=1n=1n=1 to n=2n=2n=2
ΔE=E2−E1\Delta E = E_2 - E_1ΔE=E2​−E1​ ΔE=(−0.5445×10−18)−(−2.178×10−18)\Delta E = \left(-0.5445 \times 10^{-18}\right) - \left(-2.178 \times 10^{-18}\right)ΔE=(−0.5445×10−18)−(−2.178×10−18) ΔE=1.6335×10−18 J\Delta E = 1.6335 \times 10^{-18} \text{ J}ΔE=1.6335×10−18 J
  1. Use photon energy relation

For absorption of light,

ΔE=hcλ\Delta E = \frac{hc}{\lambda}ΔE=λhc​

Hence,

λ=hcΔE\lambda = \frac{hc}{\Delta E}λ=ΔEhc​

Substitute the given values:

λ=(6.62×10−34)(3.0×108)1.6335×10−18\lambda = \frac{(6.62 \times 10^{-34})(3.0 \times 10^8)}{1.6335 \times 10^{-18}}λ=1.6335×10−18(6.62×10−34)(3.0×108)​
  1. Calculate

First, numerator:

6.62×10−34×3.0×108=19.86×10−26=1.986×10−256.62 \times 10^{-34} \times 3.0 \times 10^8 = 19.86 \times 10^{-26} = 1.986 \times 10^{-25}6.62×10−34×3.0×108=19.86×10−26=1.986×10−25

Now,

λ=1.986×10−251.6335×10−18\lambda = \frac{1.986 \times 10^{-25}}{1.6335 \times 10^{-18}}λ=1.6335×10−181.986×10−25​ λ=(1.9861.6335)×10−7\lambda = \left(\frac{1.986}{1.6335}\right) \times 10^{-7}λ=(1.63351.986​)×10−7 λ≈1.216×10−7 m\lambda \approx 1.216 \times 10^{-7} \text{ m}λ≈1.216×10−7 m

This matches closely with:

1.214×10−7 m1.214 \times 10^{-7} \text{ m}1.214×10−7 m
  1. Option check
  • A: 2.816×10−72.816 \times 10^{-7}2.816×10−7 m ❌
  • B: 6.500×10−76.500 \times 10^{-7}6.500×10−7 m ❌
  • C: 8.500×10−78.500 \times 10^{-7}8.500×10−7 m ❌
  • D: 1.214×10−71.214 \times 10^{-7}1.214×10−7 m ✅

Therefore, the correct answer is Option D.

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