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When two deuterium nuclei fuse together to form a tritium nuclei, we get a [CPMT 2000]

Neutron

Deutron

$α$ -particle

Proton

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Explanation

${}_{1}H_{2} + {}_{1}H_{2} \to {}_{1}H_{3} + {}_{1}H_{1}$

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The most penetrating radiation out of the following is [1997]

$γ$ -rays

$α$ -particles

$β$ -rays

X-rays

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Explanation

The penetrating power of radiation is directly proportional to the energy of its photon.

Energy of photon = $\frac{h c}{λ}$ $α \frac{1}{λ}$ (wavelength)

$∴$ Penetrating power $α \frac{1}{λ}$ (wavelength)

Since, $λ$ is minimum for y-rays, so penetrating power is maximum for y-rays.

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In any fission process the ratio $\frac{m a s s o f f i s s i o n p r o d u c t s}{m a s s o f p a r e n t n u c l e u s}$ is [2005]

less than 1

greater than 1

equal to 1

depends on the mass of parent nucleus

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Explanation

In fission process, when a parent nucleus breaks into daughter products, then some mass is lost in the form of energy. Thus, mass of fission products < mass of parent nucleus

$\Rightarrow \frac{M a s s o f f i s s i o n p r o d u c t s}{M a s s o f p a r e n t n u c l e u s} < 1$

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The nucleus ${}_{48}C d^{115}$ , after two successive $β$ -decay will give [1988]

${}_{46}P a^{115}$

${}_{49}I n^{114}$

${}_{50}S n^{113}$

${}_{50}S n^{115}$

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Explanation

When a parent nucleus emits a $β$ -particle mass number remains same because mass of electron is negligibly low. Atomic number is increased by one. The nucleus ${}_{48}C d^{115}$ after two successive $β$ -decays will given ${}_{50}S n^{115}$ .

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In nuclear fission process, energy is released because [2001]

mass of products is more than mass of nucleus.

total binding energy of products formed due to nuclear fission is more than that the parent fissionable material.

total binding energy of products formed due to nuclear fission is less than parent fissionable material.

mass of some particles is converted into energy.

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Explanation

In a nuclear process, energy is released if binding energy per nucleon of the daughter products gets increased. In nuclear fission reaction, total binding energy of products formed due to nuclear fission is more than the parent fissionable material.

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The average binding energy of a nucleon inside an atomic nucleus is about [1989]

8 MeV

8 eV

8 J

8 erg

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Explanation

Average binding energy per nucleon of a nucleus is the average enegy we have to spend to remove a nucleon from the nucleus to infinite distance. It is given by total binding energy divided by the mass number of the range A = 30 to A = 120 corresponding to average binding energy per nucleon = 8 MeV. The peak value of the maximum is 8 MeV / N for ${}_{26}F e^{56}$ .

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A nucleus with Z = 92 emits the following in a sequence:

$α, β^{-}, β^{-}, α, α, α, α, α, β^{-}, β^{-}, α, β^{+}, β^{+}, α$ . The Z of the resulting nucleus is [AIEEE 2003]

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Explanation

$Z_{R e s u l t i n g n u c l e u s} = 92 - 8 \times 2 + 4 \times 1 - 2 \times 1 = 78$

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$Two nucleons are at a separation of 1 fm . The net force between them is F_{1} if both are neutrons, F_{2} if both are protons and F_{3} if one is proton and other is a neutron . Hence$

$F_{1} > F_{2} > F_{3}$

$F_{2} > F_{1} > F_{3}$

$F_{1} = F_{3} > F_{2}$

$F_{1} = F_{2} > F_{3}$

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Explanation

Nuclear force independent of charge but if the electrostatic force is considerable then force between neutrons and neutron-proton pair is greater because electrostatic repulsion is there among the protons

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$A nucleus {}_{Z}^{A}X emits one α - particle and one β - particle . The resulting nucleus is$

${}_{Z - 1}^{A - 2}X$

${}_{Z - 1}^{A - 4}X$

${}_{Z}^{A - 4}X$

${}_{Z - 2}^{A - 4}X$

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Explanation

$(2) {}_{Z}X_{4} \overset{α}{\to} {}_{Z - 2}X'^{(A - 4)} \overset{β}{\to} {}_{Z - 1}X^{(A - 4)}$

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The ratio of the radius of the nucleus to the radius of an atom is of the order of :

$10^{- 5}$

$10^{- 7}$

$10^{- 2}$

$10 °$

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Explanation

The radius of the nucleus is the order of fermi and atom is the order of angstrom.

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