Nuclei Test

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Nuclei Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

If $$M_0$$ is the mass of an oxygen isotope $$_8O^{17}, M_p$$ and $$M_n$$ are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is

A
$$(M_O-8M_p)c^2$$

B
$$(M_O-8M_p-9M_n)c^2$$

C
$$M_Oc^2$$

D
$$(M_O-17M_n)c^2$$

Solution

Number of proton in Oxygen isotope, $$Z =8$$
Number of neutron, $$N = 17 - 8 = 9$$
The nuclear reaction: $$8p$$ $$ + $$ $$9 n $$ $$\rightarrow$$ $$_8O^{17}$$
Nuclear binding energy of the isotope, $$B.E = (M_{product} - M_{Reactant}) c^2$$
$$\Rightarrow $$ $$B.E = (M_{O} - 8M_p - 9M_n) c^2$$
Question 2
1 / -0

Hydrogen has________ isotopes.

A
2

B
3

C
4

D
5

Solution

Hydrogen has 3 isotopes namely. protium $$_1^1H$$, deuterium $$_1^2H$$ and tritium $$_1^3H$$
Question 3
1 / -0

Mass number is denoted by:

A
D

B
S

C
A

D
Z

Solution

The mass number (A) is the number of nucleons, which is the total number of protons and neutrons in the nucleus of an atom.
Question 4
1 / -0

One atomic mass is equal to

A
$$1.66\times10^{-27}$$ g

B
$$1.66\times10^{-24}$$ g

C
$$1.66\times10^{-23}$$ g

D
$$1.66\times10^{-25}$$ g

Solution

One atomic mass unit is equal to $$1.66 \times 10^{-24} g.$$ It is equal to 1/12th of mass of atom of Carbon-12. It is abbreviated as $$\text{u.}$$
Question 5
1 / -0

If $${}_{92}^{238}U$$ emits $$8$$ $$\alpha$$-particles and $$6$$ $$\beta$$-particles, then the resulting nucleus is

A
$${}_{82}^{206}U$$

B
$${}_{82}^{206}Pb$$

C
$${}_{82}^{210}Pb$$

D
$${}_{82}^{214}Pb$$

Solution

After emitting 1 $$\alpha$$ particle, the atomic number decreases by 2 and mass number decreases by 4.
After emitting 1 $$\beta$$ particle, the atomic number increases by 1 and mass number is unaffected.
So, in this situation, the mass number of the daughter nucleus is $$238-8\times 4=206$$ and the atomic number is $$92-8\times 2+6=82$$.
Thus, the answer is $$_{ 82 }^{ 206 }{ Pb }$$.
Question 6
1 / -0

$${m}_{p}$$ and $${m}_{n}$$ are masses of proton and neutron respectively. An element of mass $$M$$ has $$Z$$ protons and $$N$$ neutrons then:

A
$$M> Z{m}_{p}+N{m}_{n}$$

B
$$M=Z{m}_{p}+N{m}_{n}$$

C
$$M< Z{m}_{p}+N{m}_{n}$$

D
$$M$$ may be greater than less than or equal to $$Z{m}_{p}+N{m}_{n}$$, depending on nature of element

Solution

When a nucleus is formed, then the mass of nucleus is slightly less than the sum of the mass of $$Z$$ protons and $$N$$ neutrons
i.e. $$M< (Z{m}_{p}+N{m}_{n})$$
Question 7
1 / -0

If $${M}_{0}$$ is the mass of an oxygen isotope $$_{8}{O}^{17}, {M}_{p}$$ and $${M}_{n}$$ are the masses of a proton and a neutron, respectively, then the nuclear binding energy of the isotope is

A
$$\left({M}_{o} - 8{M}_{p}\right) {c}^{2}$$

B
$$\left({M}_{o} - 8{M}_{p} - 9{M}_{n}\right) {c}^{2}$$

C
$${M}_{o}{c}^{2}$$

D
$$\left({M}_{o} - 17{M}_{n}\right) {c}^{2}$$

Solution

$$\begin{aligned}M_{0} &=\text{mass of $_8O^{17}$}\\M_{p} &=\text { Mass of proton } \mathrm{} \\M_{n} &=\text { Mass of neutron } \mathrm{}_{}\end{aligned}$$

$$\text { Binding energy }=\text { mass defect }(\Delta m)\times c^{2}$$

$$\begin{array}{l}\Delta m=\left(\text { Mass of Oxygen $-$ total mass of }{\text {protons }}\right.- \text {total mass of}\text { neutrons } \\\text { Binding energy }=\left(M_{0}-8M_p -9M_n\right) c^{2}\end{array}$$
Question 8
1 / -0

When two light nuclei fuse to form a relatively heavier nucleus, the specific binding energy of the product nucleus is:

A
lower than that of the reacting nuclei

B
equal to that of the reacting nuclei

C
greater than that of the reacting nuclei

D
equal to exactly half of either of the reacting nuclei

Solution

The binding energy of the product nucleus will be greater than that of the reacting nuclei, because when two light nuclei fuse to form relatively heavier nucleus, energy is released. And the higher the binding energy, the more stable the nucleus is.
Question 9
1 / -0

An alpha particle $$(_{ }^{ 4 }{ He })$$ has a mass $$4.00300\ amu$$. A proton has a mass $$1.00783$$ amu and a neutron has a mass $$1.00867\ amu$$ respectively. The binding energy of alpha estimated from these data is closest to:

A
$$27.9\ MeV$$

B
$$22.3\ MeV$$

C
$$35.0\ MeV$$

D
$$20.4\ MeV$$

Solution

The mass defect, $$\Delta m=2\left( { m }_{ p }+{ m }_{ n } \right) -{ m }_{ He }$$
$$=2(1.00783+1.00867)-4.00300$$
$$=0.300\ amu$$
So, the binding energy $$E=\Delta m{ c }^{ 2 }$$$$=0.03\times 931MeV=27.9\ MeV\quad $$
Question 10
1 / -0

When three $$\alpha-particles$$ combined to from a $$^{12}C$$ nucleus, the mass defect is
(atomic mass of $$_{2}He^{4}$$ is $$4.002603\ u$$)

A
$$0.007809\ u$$

B
$$0.002603\ u$$

C
$$4.002603\ u$$

D
$$0.5\ u$$

Solution

Mass defect, $$\triangle =$$ Total mass of $$\alpha-particles -$$ mass of $$^{12}C$$ nucleus
$$= 3\times 4.002603 - 12$$
$$= 12.007809 - 12$$
$$= 0.007809\ unit$$

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Nuclei Test - 38

Result

Nuclei Test - 38

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Question 1

$$(M_O-8M_p)c^2$$

$$(M_O-8M_p-9M_n)c^2$$

$$M_Oc^2$$

$$(M_O-17M_n)c^2$$

Solution

Question 2

2

3

4

5

Solution

Question 3

D

S

A

Z

Solution

Question 4

$$1.66\times10^{-27}$$ g

$$1.66\times10^{-24}$$ g

$$1.66\times10^{-23}$$ g

$$1.66\times10^{-25}$$ g

Solution

Question 5

$${}_{82}^{206}U$$

$${}_{82}^{206}Pb$$

$${}_{82}^{210}Pb$$

$${}_{82}^{214}Pb$$

Solution

Question 6

$$M> Z{m}_{p}+N{m}_{n}$$

$$M=Z{m}_{p}+N{m}_{n}$$

$$M< Z{m}_{p}+N{m}_{n}$$

$$M$$ may be greater than less than or equal to $$Z{m}_{p}+N{m}_{n}$$, depending on nature of element

Solution

Question 7

$$\left({M}_{o} - 8{M}_{p}\right) {c}^{2}$$

$$\left({M}_{o} - 8{M}_{p} - 9{M}_{n}\right) {c}^{2}$$

$${M}_{o}{c}^{2}$$

$$\left({M}_{o} - 17{M}_{n}\right) {c}^{2}$$

Solution

Question 8

lower than that of the reacting nuclei

equal to that of the reacting nuclei

greater than that of the reacting nuclei

equal to exactly half of either of the reacting nuclei

Solution

Question 9

$$27.9\ MeV$$

$$22.3\ MeV$$

$$35.0\ MeV$$

$$20.4\ MeV$$

Solution

Question 10

$$0.007809\ u$$

$$0.002603\ u$$

$$4.002603\ u$$

$$0.5\ u$$

Solution

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