Physics Nuclei
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A radioactive nucleus undergoes spontaneous decay in the sequence , where is the atomic number of element . The possible decay particles in the sequence are:
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A radioactive nucleus undergoes spontaneous decay in the sequence , where is the atomic number of element . The possible decay particles in the sequence are:
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decreases atomic number by 1
decreases atomic number by 2
increases atomic number by 1
In the given nuclear reaction, the elements X is:
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In the given nuclear reaction, the elements X is:
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The decay of a proton to neutron is :
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The decay of a proton to neutron is :
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Always possible as it is associated only with β? decay
A radioactive sample disintegrates via two independent decay processes having half lives T?/?¹? and T?/?²? respectively. The effective half-life, T?/? of the nuclei is:
A radioactive sample disintegrates via two independent decay processes having half lives T?/?¹? and T?/?²? respectively. The effective half-life, T?/? of the nuclei is:
From Radioactive Decay Law,
dN/dt = λ? N + λ? N = λ_eff N
⇒ λ_eff = λ? + λ? ⇒ ln (2)/T = ln (2)/T? + ln (2)/T? ⇒ T = (T? ) / (T? + T? )
(where T, T? , and T? are half-lives)
Given the masses of various atomic particles m? = 1.0072u, m? = 1.0087u, m? = 0.000548u, m? = 0, m? = 2.0141u, where p = proton, n = neutron, e = electron, v = antineutrino and d = deuteron. Which of the following process is allowed by momentum and energy conservation?
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Given the masses of various atomic particles m? = 1.0072u, m? = 1.0087u, m? = 0.000548u, m? = 0, m? = 2.0141u, where p = proton, n = neutron, e = electron, v = antineutrino and d = deuteron. Which of the following process is allowed by momentum and energy conservation?
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Total mass of reactant should be greater than that of product.
This condition is only fulfilled in case- 3
In a radioactive material, fraction of active material remaining after time t is 9/16. The fraction that was remaining after t/2 is
In a radioactive material, fraction of active material remaining after time t is 9/16. The fraction that was remaining after t/2 is
The law of radioactive decay is N = N? e? λt, where N is the amount remaining at time t.
Given that at time t, N/N? = 9/16.
So, 9/16 = e? λt
At time t/2, the fraction remaining will be N'/N?
N' = N? e? λ ( t/2 ) = N? (e? λt)¹/²
Substituting the value of e? λt:
N' = N? (9/16)¹/² = N? (3/4)
The fraction re
In a reactor, 2 kg of ₉₂U²³⁵ fuel is fully used up in 30 days. The energy released per fission is 200MeV. Given that the Avogadro number, N = 6.023 × 10²⁶ per kilo mole and 1eV = 1.6 × 10⁻¹⁹ J. The power output of the reactor is close to:
(A) 60MW
In a reactor, 2 kg of ₉₂U²³⁵ fuel is fully used up in 30 days. The energy released per fission is 200MeV. Given that the Avogadro number, N = 6.023 × 10²⁶ per kilo mole and 1eV = 1.6 × 10⁻¹⁹ J. The power output of the reactor is close to:
(A) 60MW
P = E/t
= 2/235 × (6.023×10²? ×200×1.6×10? ¹? )/ (30×24×60×60) = 60W
You are given that Mass of ?Li = 7.0160u, Mass of ?He = 4.0026u and ¹H = 1.0079u. When 20 g of ?Li is converted into ?He by proton capture, the energy liberated, (in kWh), is [Mass of nucleon = 1GeV/c²]
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You are given that Mass of ?Li = 7.0160u, Mass of ?He = 4.0026u and ¹H = 1.0079u. When 20 g of ?Li is converted into ?He by proton capture, the energy liberated, (in kWh), is [Mass of nucleon = 1GeV/c²]
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Kindly consider the following figure