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86A^222rarr84B^210. In this reaction, ho...

`_86A^222rarr_84B^210`. In this reaction, how many `alpha` and `beta` particles are emitted?

A

`6alpha , 3 beta`

B

`3 alpha , 4 beta`

C

`4 alpha , 3 beta`

D

`3 alpha, 6 beta`

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The correct Answer is:
To solve the problem of how many alpha and beta particles are emitted in the reaction `_86A^222 → _84B^210`, we will follow these steps: ### Step 1: Identify the Initial and Final States - The initial element A has: - Mass number (A) = 222 - Atomic number (Z) = 86 - The final element B has: - Mass number (A) = 210 - Atomic number (Z) = 84 ### Step 2: Calculate the Change in Mass Number - The change in mass number (ΔA) can be calculated as: \[ \Delta A = A_{initial} - A_{final} = 222 - 210 = 12 \] ### Step 3: Determine the Number of Alpha Particles Emitted - Each alpha particle emission decreases the mass number by 4. Therefore, the number of alpha particles (nα) emitted can be calculated as: \[ n_{\alpha} = \frac{\Delta A}{4} = \frac{12}{4} = 3 \] ### Step 4: Calculate the Change in Atomic Number - The change in atomic number (ΔZ) can be calculated as: \[ \Delta Z = Z_{initial} - Z_{final} = 86 - 84 = 2 \] ### Step 5: Relate Alpha and Beta Particle Emissions - Each alpha particle emission decreases the atomic number by 2. Therefore, the contribution of alpha particles to the atomic number change is: \[ \text{Change due to alpha particles} = 2 \times n_{\alpha} = 2 \times 3 = 6 \] - Since the atomic number decreases by 6 due to alpha emissions, we need to find how many beta particles (nβ) are emitted to adjust the atomic number back to 84. Each beta particle increases the atomic number by 1. Thus, we can set up the equation: \[ Z_{final} = Z_{initial} - 2n_{\alpha} + n_{\beta} \] Plugging in the values: \[ 84 = 86 - 6 + n_{\beta} \] ### Step 6: Solve for the Number of Beta Particles - Rearranging the equation gives: \[ n_{\beta} = 84 - 80 = 4 \] ### Conclusion - The total emissions are: - Number of alpha particles emitted (nα) = 3 - Number of beta particles emitted (nβ) = 4 Thus, the final answer is: - **3 alpha particles and 4 beta particles are emitted.** ---

To solve the problem of how many alpha and beta particles are emitted in the reaction `_86A^222 → _84B^210`, we will follow these steps: ### Step 1: Identify the Initial and Final States - The initial element A has: - Mass number (A) = 222 - Atomic number (Z) = 86 - The final element B has: ...
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DC PANDEY ENGLISH-MODERN PHYSICS - 1-Level 1 Objective
  1. According to Einstein's photoelectric equation, the graph of kinetic e...

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  2. The ratio of the speed of the electron in the first Bohr orbit of hydr...

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  3. 86A^222rarr84B^210. In this reaction, how many alpha and beta particl...

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  4. An X-ray tube is operated at 20 kV. The cut off wavelength is

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  5. An X-ray tube is opearted at 18 kV. The maximum velocity of electron s...

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  6. what is the ratio of de-Broglie wavelength of electron in the second a...

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  7. The energy of a hydrogen like atom (or ion) in its ground state is -12...

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  8. The operating potential in an x-ray tube is increased by 2%. The perce...

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  9. The energy of an atom or ion in the first excited state is -13.6 eV. I...

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  10. in order that the short wavelength limit of the continuous X-ray spec...

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  11. The momentum of an x-ray photon with lambda = 0.5 Å is

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  12. The work function of a substance is 1.6 ev. The longest wavelength of ...

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  13. Find the binding anergy of an electron in the ground state of a hydrog...

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  14. What is the energy of a hydrogen atom in the first excited state if th...

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  15. Light of wavelength 330nm falling on a piece of metal ejects electrons...

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  16. Maximum kinetic energy of a photoelectron is E when the wavelength of ...

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  17. if the frequency fo Ka X-ray emitted from the element with atomic numb...

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  18. According to Moseley's law the ratio of the slope of graph between sqr...

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  19. If the electron in an hydrogen atom jumps from an orbit with level n(f...

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  20. A potential of 10000 V is applied across an x-ray tube. Find the ratio...

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