Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Atoms and Nuclei question

2023 · 13 Apr · Shift 1 · Q55
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Atoms and Nuclei
  5. /2023 · 13 Apr · Shift 1 · Q55

Atoms and Nuclei question

2023 · 13 Apr · Shift 1 · Q55

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
92238A→90234B+24D+Q_{92}^{238}A \to _{90}^{234}B + _2^4D + Q92238​A→90234​B+24​D+Q In the given nuclear reaction, the approximate amount of energy released will be: [Given, mass of 92238 A=238.05079×931.5 MeV/c2,{ }_{92}^{238} \mathrm{~A}=238.05079 \times 931.5 ~\mathrm{MeV} / \mathrm{c}^{2},92238​ A=238.05079×931.5 MeV/c2, mass of 90234B=234⋅04363×931⋅5 MeV/c2,{ }_{90}^{234} B=234 \cdot 04363 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2},90234​B=234⋅04363×931⋅5 MeV/c2, mass of 24D=4⋅00260×931⋅5 MeV/c2]\left.{ }_{2}^{4} D=4 \cdot 00260 \times 931 \cdot 5 ~\mathrm{MeV} / \mathrm{c}^{2}\right]24​D=4⋅00260×931⋅5 MeV/c2]
  1. A
    2.12 MeV
  2. B
    4.25 MeV
  3. C
    3.82 MeV
  4. D
    5.9 MeV
View written solutionFree

Correct answer: B

  1. Use the QQQ-value formula

For the nuclear reaction

92238A→90234B+24D+Q,{}^{238}_{92}A \to {}^{234}_{90}B + {}^{4}_{2}D + Q,92238​A→90234​B+24​D+Q,

the energy released is

Q=[m(A)−m(B)−m(D)]c2.Q = \left[m(A)-m(B)-m(D)\right]c^2.Q=[m(A)−m(B)−m(D)]c2.

Since the masses are already given in MeV/c2/c^2/c2, we can directly subtract and then multiply by c2c^2c2, which numerically gives the result in MeV.

  1. Substitute the given masses

Given:

m(A)=238.05079×931.5 MeV/c2,m(A)=238.05079\times 931.5\,\text{MeV}/c^2,m(A)=238.05079×931.5MeV/c2, m(B)=234.04363×931.5 MeV/c2,m(B)=234.04363\times 931.5\,\text{MeV}/c^2,m(B)=234.04363×931.5MeV/c2, m(D)=4.00260×931.5 MeV/c2.m(D)=4.00260\times 931.5\,\text{MeV}/c^2.m(D)=4.00260×931.5MeV/c2.

So,

Q=[(238.05079−234.04363−4.00260)×931.5]MeV.Q = \left[(238.05079-234.04363-4.00260)\times 931.5\right]\text{MeV}.Q=[(238.05079−234.04363−4.00260)×931.5]MeV.
  1. Calculate mass defect
Δm=238.05079−234.04363−4.00260\Delta m = 238.05079 - 234.04363 - 4.00260Δm=238.05079−234.04363−4.00260 Δm=238.05079−238.04623=0.00456\Delta m = 238.05079 - 238.04623 = 0.00456Δm=238.05079−238.04623=0.00456
  1. Convert mass defect into energy
Q=0.00456×931.5 MeVQ = 0.00456 \times 931.5\,\text{MeV}Q=0.00456×931.5MeV Q≈4.24764 MeVQ \approx 4.24764\,\text{MeV}Q≈4.24764MeV

Hence,

Q≈4.25 MeV.Q \approx 4.25\,\text{MeV}.Q≈4.25MeV.
  1. Match with the options

The closest option is:

B: 4.25 MeV4.25\,\text{MeV}4.25MeV

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer is also B, so they agree.

PreviousNext

More from Atoms and Nuclei

  • The radius of 2nd  orbit of He+ of Bohr's model is r1​ and that of fourth orbit of Be3+ is represented as r2​. Now the ratio r1​r2​​ is x:1. The value of x is ​…2023 · Numerical
  • As per given figure A,B and C are the first, second and third excited energy levels of hydrogen atom respectively. If the ratio of the two wavelengths ( i.e. λ2​λ1​​) is 4n7​… Includes diagram2023 · Numerical
  • A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4 × 10 −15 eVs) :2023 · MCQ
  • The energy released per fission of nucleus of 240 X is 200 MeV. The energy released if all the atoms in 120g of pure 240 X undergo fission is ​× 10 25 MeV. (Given NA​=6×1023)2023 · Numerical
  • The wavelength of the radiation emitted is λ0​ when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen…2023 · Numerical
  • The energy levels of an atom is shown in figure. Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm? Given (h = 6.62 × 10 −34 Js) Includes diagram2023 · MCQ
  • Speed of an electron in Bohr's 7th  orbit for Hydrogen atom is 3.6×106 m/s. The corresponding speed of the electron in 3rd  orbit, in m/s is :2023 · MCQ
  • For hydrogen atom, λ1​ and λ2​ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of λ1​ and λ2​ is 32x​. The value of x is ​… Includes diagram2023 · Numerical