Nuclei Test

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Nuclei Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

For mass defect of $$0.3\%$$, the binding energy of $$1kg$$ material is :

A
$$2.7\times 10^{14}erg$$

B
$$2.7\times 10^{14}J$$

C
$$2.7\times 10^{-14}erg$$

D
$$2.7\times 10^{-14}J$$

Solution

$$\Delta m=0.3 \% \text { of } 1 \mathrm{~kg} \\$$
$$\Delta m=\frac{0.3}{100} \\$$
$$\Delta m=3 \times 10^{-3} \mathrm{~kg} \\$$
$$E=(\Delta m) c^{2} \\$$
$$E=3 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2} \\$$
$$E=3 \times 9 \times 10^{-3} \times 10^{16} \mathrm{~J} \\$$
$${[B] E=2.7 \times 10^{14} \mathrm{~J}}$$
Question 2
1 / -0

Write the S.I. unit of activity.

A
$$Becquerel$$

B
$$Henry$$

C
$$Ohm$$

D
$$Mendel$$

Solution

$$S.I.$$ Unit of activity : It is the count rate of radioactivity.
It's $$S.I$$ Unit is decay per second $$=1\ Bq$$
where $$Bq=Becquerel (\because 1\ Bq=1\ decay\ per second)$$
Question 3
1 / -0

The wavelength of $$K _ { \alpha }$$ line for an element of atomic number $$43$$ is $$\lambda.$$ Then the wavelength of $$K _ { \alpha }$$ line for an element of atomic number $$29$$ is

A
$$\dfrac { 43 } { 29 } \lambda$$

B
$$\dfrac { 42 } { 28 } \lambda$$

C
$$\dfrac { 9 } { 4 } \lambda$$

D
$$\dfrac { 4 } { 9 } \lambda$$

Solution

As we know,
$$\lambda \propto \dfrac{1}{(Z-1)^2} \Rightarrow \dfrac{\lambda_2}{\lambda_1}=\left(\dfrac{Z_1-1}{Z_2-1}\right)^2$$
$$\Rightarrow \dfrac{\lambda_2}{\lambda}=\left(\dfrac{43-1}{29-1}\right)^2=\left(\dfrac{42}{28}\right)^2 \Rightarrow \lambda_2=\dfrac{9}{4} \lambda$$
Question 4
1 / -0

The nuclear radius is given by $$R = r_0A^{1/3}$$ , where $$r_0$$ is constant and A is the atomic mass number.
Then :

A
The nuclear mass density of $$U^{238}$$ is twice that of $$Sn^{119}$$.

B
The nuclear mass density of $$U^{238}$$ is thrice that of $$Sn^{119}$$

C
The nuclear mass density of $$U^{238}$$ is the same as that of $$Sn^{119}$$

D
The nuclear mass density of $$U^{238}$$ is half that of $$Sn^{119}$$

Solution

$$|C|$$
$$R=r_0A^{1/3}$$
$$R^3=r^3_0A$$
$$Density=\dfrac{Mass}{Volume}=\dfrac{Am_p}{\dfrac{4}{3}\pi R^3}=\dfrac{Am_p}{\dfrac{4}{3}\pi r^3_0A}=\dfrac{3m_p}{4\pi r^3_0}$$
Hence mass density does not depend on mass (A) or atomic number (z)
Question 5
1 / -0

Protons of energy $$2$$eV and $$2.5$$eV successively illuminate a metal whose work function is $$0.5$$eV. The ratio of maximum speed of emitted electron is _______.

A
$$\sqrt{3}:2$$

B
$$2:1$$

C
$$1:2$$

D
$$2:\sqrt{3}$$

Solution

$$K.E_{max}=hv-\phi$$
$$\dfrac{1}{2}mv^2=hv-\phi$$
$$\dfrac{v^2_1}{v^2_2}=\dfrac{2-0.5}{2.5-0.5}=\dfrac{1.5}{2}=\dfrac{3}{4}$$
$$\dfrac{v_1}{v_2}=\dfrac{\sqrt{3}}{2}$$.
Question 6
1 / -0

The notation for an isotope of sodium $$^{23}_{11}Na$$. Which row gives the composition of a neutral atom of this isotope of sodium?

number of protons number of neutrons number of electrons
A $$11$$ $$12$$ $$11$$
B $$11$$ $$12$$ $$12$$
C $$11$$ $$23$$ $$11$$
D $$12$$ $$11$$ $$12$$

A
A

B
B

C
C

D
D

Solution

In $$^{23}_{11}Na$$
Atomic No. , $$A= 11$$
Mass No. ,$$M= 23$$

Number of protons= $$A= 11$$
As it its not ionized
Number of electrons $$=A=11$$

Number of neutrons,$$=M-A=23-11=12$$
Question 7
1 / -0

Molecular mass of dry air is ________________.

A
less than moist air

B
greater than moist air

C
equal to moist air

D
may be greater or less than moist air

Solution

The molecular mass of dry ice is greater than moist air.
As dry air consists of nitrogen and oxygen while moist air contains water vapour which less mass than that of oxygen and nitrogen making the mass of moist air lower.

$$M.M $$ of dry air $$\simeq 28.9\ gm$$

$$M.M $$ of moist air $$< 28.9\ gm$$
Question 8
1 / -0

Which diagram represents a nucleus of $$^3_1H$$?

A

B

C

D

Solution

Here
n$$=$$ a neutron
p$$=$$ a proton.

In $$^3_1 H$$
As its Atomic No. is $$1$$. So the number of protons is $$1$$
And its Mass no is $$3$$

No. of neutrons = Atomic No. - Mass No. $$=3-1=2$$

So follwing diagram is correct structure of $$^3_1H$$
Question 9
1 / -0

When the atomic number $$A$$ of a nucleus increases :

A
initially the neutron-proton ratio is constant

B
initially the neutron-proton per nucleon increases and then decreases

C
initially the binding energy per nucleon increases and then decreases

D
the binding energy per nucleon increases when neutron proton ratio increases

Solution
Question 10
1 / -0

$$n=p+e$$, $$n=$$ neutron, $$p=$$ proton, $$e=$$ electron
The decay equation is

A
correct

B
wrong

C
sometimes correct

D
sometimes wrong

Solution