Nuclei Test

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Nuclei Test - 31

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Question 1
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Directions For Questions

The sun is an average star of diameter of about $$1.4\times10^{5}$$ and is about $$1.5\times10^{8}$$ km distance away from earth. The average density of the Sun, which consists $$70$$ % $$H$$ and $$28$$ % $$He$$, is about $$1.4$$ times density of water. The pressure and temperatures inside the sun is very large, the temperature being around $$14$$ million $$K$$. At present high temperature thermonuclear reactions take place, converting lighter into heavier ones

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The fusion of light elements take place at about the temperatures of about

A
$$30\ ^{0}C$$

B
$$100\ ^{0}C$$

C
$$10,000\ ^{0}C$$

D
$$2\times10\ ^{0}C$$

Solution

Thermonuclear reaction, fusion of two light atomic nuclei into a single heavier nucleus by a collision of the two interacting particles at extremely high temperatures, with the consequent release of a relatively large amount of energy.

The Sun is a main-sequence star, and thus generates its energy by nuclear fusion of hydrogen nuclei into helium. In its core, the Sun fuses 620 million metric tons of hydrogen each second.
as shown in the figure that hydrogen atom are fusses to helium atom.
Question 2
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The nucleus obtained after $$\alpha$$ -emission from the nucleus $$_{y}\textrm{A}^{x}$$ is

A
$$_{y-2}\textrm{B}^{x-2}$$

B
$$_{y+2}\textrm{B}^{x+4}$$

C
$$_{y}\textrm{B}^{x}$$

D
$$_{y-2}\textrm{B}^{x-4}$$

Solution

The emission of an alpha particle from a nucleus reduces the mass number of the nucleus by $$4$$ and the atomic number by $$2$$.
Question 3
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The energy of thermal neutrons is nearly

A
$$0.25$$ $$eV$$

B
$$0.025$$ $$eV$$

C
$$200$$ $$MeV$$

D
$$0.025$$ $$Joule$$

Solution

A thermal neutron is a free neutron with a kinetic energy of about $$0.025\ eV$$ (about $$4.0\times10^{-21}J $$ or $$2.4\ MJ/kg$$, hence a speed of $$2.2\ km/s$$), which is the energy corresponding to the most probable velocity at a temperature of $$290\ K$$.
Question 4
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The particle $$X$$ in the following nuclear reaction is $$_{3}^{7}{Li}$$ + $$_{1}^{1}\textrm{H}$$ $$\longrightarrow $$ $$_{2}^{4}\textrm{He}$$ + $$X$$

A
$$a$$

B
$$n$$

C
$$e$$

D
$$p$$

Solution

Mass number of $$Li$$ is $$7$$.
So $$X$$ should have a mass number of $$4$$ and atomic number of $$2$$.
$$X$$ is a $$\alpha$$ particle.
Question 5
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Which of the following is correct?

A
There are $$78$$ neutrons in $$_{78}{Pt}^{192}$$

B
$$_{84}{Po}^{214}$$ $$\rightarrow$$ $$_{82}{Pb}^{210}$$ + $$\beta$$

C
$$_{92}{U}^{238}$$ $$\rightarrow$$ $$_{90}{Th}^{234}$$ + $$_{2}{He}^{4}$$

D
$$_{90}{Th}^{234}$$ $$\rightarrow$$ $$_{91}{Pa}^{234}$$+ $$_{2}{He}^{4}$$

Solution

After release of helium, there will be decrease in atomic number by $$2$$ and mass number by $$4$$.
Question 6
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The particle $$X$$ in the following nuclear reaction is $$_{7}^{13}\textrm{N}$$ $$\longrightarrow $$ $$_{6}^{13}\textrm{C}+$$ $$_{1}^{0}\textrm{e}$$ + $$X$$

A
$$P$$

B
$$v$$

C
$$e^{-}$$

D
$$\alpha$$

Solution

The given reaction : $$^{13}_7N\rightarrow ^{13}_6C + ^0_1e + ^A_Z X$$
Conservation of mass number : $$13 = 13 + 0 +A$$ $$\implies A=0$$
Conservation of atomic number : $$7 = 6 + 1 +Z$$ $$\implies Z=0$$
Thus, the particle $$X$$ is neutrino i.e. $$\nu$$
Question 7
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The particle $$X$$ in the following nuclear reaction is $$_{5}^{B}\textrm{10}$$ +$$_{2}^{He}\textrm{4}$$ $$\longrightarrow $$ $$_{7}^{N}\textrm{13}$$ + $$X$$

A
$$P$$

B
$$a$$

C
$$e$$

D
$$n$$

Solution

Since the atomic number remains same (i.e $$5+2=7$$), but mass number is changed, it is a neutron.
Question 8
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The temperature necessary for fusion reaction is

A
$$3\times10^{3}$$

B
$$3\times10^{6}$$

C
$$3\times10^{2}$$

D
$$3\times10^{4}$$

Solution

The prime energy producer in the Sun is the fusion of hydrogen to form helium, which occurs at a solar-core temperature at this temperature.
Question 9
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If $$_{5}\textrm{B}^{11}$$ converts into $$_{6}\textrm{C}^{11}$$, then the particle emitted in this process will be

A
electron

B
proton

C
neutron

D
positron

Solution

$$\beta ^-$$ minus decay produces an electron and electron anti neutrino.
$$\beta ^-$$ decay increases the proton by $$1$$ and reduces the neutron number by $$1$$
Hence, mass number remains same but atomic number increase by $$1.$$
Question 10
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An element X decays, first by positron emission and then two $$\alpha$$-particles are emitted in successive radioactive decay. If the product nucleus has a mass number $$229$$ and atomic number $$89$$, the mass number of atomic number of element X are

A
$$237, 93$$

B
$$237, 94$$

C
$$221, 84$$

D
$$237, 92$$

Solution

$$X \rightarrow e^+ + 2 \alpha + Y$$
$$\alpha = He^{2+}$$
Mass of $$\alpha =4$$

Mass no. of $$Y = 229$$
By conservation of mass:
Mass no. of $$X = My + 8 = 237$$

Conservation of charge:
$$\alpha = 2n + 2p$$
No. of protons in $$Y = 89$$
No. of protons in $$X= 89 + 4+1$$(positron) $$= 94$$
Hence, atomic no. of $$X = 94$$

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Re-Attempt Weekly Quiz Competition

Nuclei Test - 31

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Nuclei Test - 31

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Question 1

$$30\ ^{0}C$$

$$100\ ^{0}C$$

$$10,000\ ^{0}C$$

$$2\times10\ ^{0}C$$

Solution

Question 2

$$_{y-2}\textrm{B}^{x-2}$$

$$_{y+2}\textrm{B}^{x+4}$$

$$_{y}\textrm{B}^{x}$$

$$_{y-2}\textrm{B}^{x-4}$$

Solution

Question 3

$$0.25$$ $$eV$$

$$0.025$$ $$eV$$

$$200$$ $$MeV$$

$$0.025$$ $$Joule$$

Solution

Question 4

$$a$$

$$n$$

$$e$$

$$p$$

Solution

Question 5

There are $$78$$ neutrons in $$_{78}{Pt}^{192}$$

$$_{84}{Po}^{214}$$ $$\rightarrow$$ $$_{82}{Pb}^{210}$$ + $$\beta$$

$$_{92}{U}^{238}$$ $$\rightarrow$$ $$_{90}{Th}^{234}$$ + $$_{2}{He}^{4}$$

$$_{90}{Th}^{234}$$ $$\rightarrow$$ $$_{91}{Pa}^{234}$$+ $$_{2}{He}^{4}$$

Solution

Question 6

$$P$$

$$v$$

$$e^{-}$$

$$\alpha$$

Solution

Question 7

$$P$$

$$a$$

$$e$$

$$n$$

Solution

Question 8

$$3\times10^{3}$$

$$3\times10^{6}$$

$$3\times10^{2}$$

$$3\times10^{4}$$

Solution

Question 9

electron

proton

neutron

positron

Solution

Question 10

$$237, 93$$

$$237, 94$$

$$221, 84$$

$$237, 92$$

Solution

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