Nuclei Test

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Nuclei Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses ail its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The min. K.E. of colliding electron will be

A
$$10.2$$ eV

B
$$1.9$$ eV

C
$$12.1$$ eV

D
$$13.6$$ eV

Solution

The electron hits the hydrogen atom in ground state with energy $$K$$.
This energy goes in exciting the atom. The electron loses all its energy.
Then, the atom emits a photon by jumping back to ground state.
So, the energy of the electron is equal to the energy of the photon emitted.
Now the maximum wavelength in the Baimer series is for the transition between $$3\to 2$$ and so we get
$$\dfrac {1}{\lambda_m} =R \left (\dfrac {1}{2^2} -\dfrac {1}{3^2}\right) =R \left (\dfrac {1}{4}-\dfrac {1}{9} \right)=\dfrac {5}{36}R$$
Now, the energy of the photon emitted is
$$E=\dfrac {hc}{\lambda_m}=\dfrac {hc5R}{36}=\dfrac {6.63 \times 10^{-34}\times 3\times 10^8 \times 5\times 1.097 \times 10^7}{36}$$
$$\therefore \ E=3.03\times 10^{-19}\ J$$
$$\therefore \ E=\dfrac {3.03\times 10^{-19}}{1.6\times 10^{-19}}=1.9\ eV$$
Hence, the correct option is $$(B)$$
Question 2
1 / -0

P and Q are two elements which form $${ P }_{ 2 }{ Q }_{ 3 }$$ and $${ PQ }_{ 2 }$$. If 0.15 mole of $${ P }_{ 2 }{ Q }_{ 3 }$$ weight 15.9 g and 0.15mole of $${ PQ }_{ 2 }$$ weight 9.3 g. what are atomic weights of P and Q respectively?

A
18 and 26

B
26 and 26

C
26 and 18

D
18 and 18

Solution

Using- no. of mole $$=\cfrac{\text {weight}}{\text{molecular weight}}$$

$$P_2Q_3 \Rightarrow $$ $$2P + 3Q =15.9/0.15= 106$$....(1)

$$PQ_2 \Rightarrow $$ $$P + 2Q = 9.3/0.15=62$$........(2)

On solving eqn. (1) & (2) we get-

$$P =26, Q = 18$$
Question 3
1 / -0

$$_{92}U^{238}$$ on absorbing a neutron goes over to $$_{92}U^{239}$$. This nucleus emits an electron to go over to neptunium which on further emitting an electron goes over to plutonium. The plutonium nucleus can be expressed as:

A
$$_{94}Pu^{239}$$

B
$$_{92}Pu^{239}$$

C
$$_{93}Pu^{240}$$

D
$$_{92}Pu^{240}$$

Solution

$$\text{Given $-*_{92} U^{238}$ on absorbing neutron $\longrightarrow _{92} U^{239}$}$$
$$\text{Now the nucleus -}$$
$$_{92}{U^{239}} \quad \underline{\text { emits an } e^{-}} \quad _{93} N p^{239}$$

$$\begin{array}{l}\text { On further emitting an electron, } \\\qquad _{93}Np^{239} \quad \underline{\text { emits an } e^{-}} \quad _{94}Pu^{239} \\\text { Since, the emission of an } e^{-}\text{increases the atomic number } \\\text { by (1) and no change in the atomic mass. } \\\Rightarrow \text { opt (a) is correct. }\end{array}$$
Question 4
1 / -0

A nucleus $$_Z{X}^A$$ emits $$9 \alpha$$-particles and $$5p$$ particle. The ration of total protons and neutrons in the final nucleus is:-

A
$$\dfrac{(Z - 13)}{(A - Z - 23)}$$

B
$$\dfrac{(Z - 18)}{(A - 36)}$$

C
$$\dfrac{(Z - 23)}{(A - Z-8)}$$

D
$$\dfrac{(Z - 13)}{(A - Z - 13)}$$

Solution

Hence, option $$C$$ is correct answer.
Question 5
1 / -0

If $$_{ a }^{ b }{ X }$$ emits a positron, two $$\alpha $$ and two $$\beta^- $$ and in last one $$\alpha $$ is also emitted and converts in $$_{ d }^{ c }{ Y }$$, correct relation:

A
$$c=b-12, d=a-5$$

B
$$a=c-8, d=b-1$$

C
$$a=c-6, d=b-0$$

D
$$a=c-4, a=b-2$$

Solution
Question 6
1 / -0

If the binding energy per nucleon in $$_{3}^{7}Li$$ and $$_{2}^{4}He$$ nuclei are $$5.60\ MeV$$ and $$7.06\ MeV$$ respectively, then in the reaction : $$p + _{3}^{7}Li \rightarrow 2_{2}^{4}He$$ energy of proton must be

A
$$28.24\ MeV$$

B
$$17.28\ MeV$$

C
$$1.46\ MeV$$

D
$$39.2\ MeV$$

Solution

$$\begin{array}{l}\text { Given }-B E \text { of }_{3}^{7} L_{i}=5.6\mathrm{MeV} , B E \text { of }_{2}^{4}\mathrm{He}=7.06\mathrm{MeV} \\p+\frac{7}{3}\mathrm{Li}\rightarrow 2 \cdot \frac{4}{2} \mathrm{He}\end{array}$$

$$\begin{aligned}\text { Energy of proton } &=2 \cdot B E \text { of }\mathrm{He}-\mathrm{BE} \text { of Li } \\&=2 \cdot[4 \times 7.06]-[7\times 5.60] \\&=17.28 \mathrm{MeV}\end{aligned}$$
Question 7
1 / -0

For a nucleus to be stable, the correct relation between neutron number N and proton number Z is

A
N>Z

B
N=z

C
N<Z

D
$$N \geqslant Z$$

Solution
Question 8
1 / -0

Let $$u$$ denote one atomic mass unit. One atom of an element of mass number $$A$$ has mass exactly equal to $$Au$$

A
for any value of $$A$$

B
only for $$A=1$$

C
only for $$A=12$$

D
for any value of $$A$$ provided the atom is stable

Solution
Question 9
1 / -0

If half-life of a radioactive substance is $$60$$ minutes, then the percentage decay in $$4$$ hours is?

A
$$50\%$$

B
$$71\%$$

C
$$85\%$$

D
$$93.7\%$$

Solution
Question 10
1 / -0

In a hydrogen atom, the binding energy of the electron in the ground state is $$E_{1}.$$ Then the frequency of revolution of nth electron in the nth orbits is

A
$$\frac{2E_{1}}{nh}$$

B
$$\frac{2E_{1}n^{3}}{h}$$

C
$$\sqrt{\frac{2mE_{1}}{n^{3}h}}$$

D
$$\frac{2E_{1}n^{2}}{h}$$

Solution

$$\begin{array}{l}\text{B.E of electron= K.E of electron (in that orbit)}\\\dfrac{1}{2}mv^2=E_1-(i)\\\therefore \text{angular momentum -} mvr = \dfrac{nh}{2 \pi}-(ii)\\\text{Now from (i) divided by (ii)}\\\dfrac{v}{2r}= \dfrac{E_1 \times 2 \pi}{nh}\\\dfrac{v}{2 \pi r}=\dfrac{2E_1}{nh}\\\implies f=\dfrac{2E_1}{nh}\end{array}$$