Blog Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

● ● ● ● ●

The velocity of an electron in the second orbit of sodium atom (atomic number = 11) is v. The velocity of an electron in its fifth orbit will be

$\frac{22}{5} v$

$\frac{5}{2} v$

$\frac{2}{5} v$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) $v_{n} \propto \frac{1}{n} \Rightarrow \frac{v_{5}}{v_{2}} = \frac{2}{5} \Rightarrow v_{5} = \frac{2}{5} v_{2} = \frac{2}{5} v$

● ● ● ● ●

The absorption transitions between the first and the fourth energy states of hydrogen atom are 3. The emission transitions between these states will be

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) By using $N_{E} = \frac{n (n - 1)}{2} \Rightarrow N_{E} = \frac{4 (4 - 1)}{2} = 6$

● ● ● ● ●

The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is

$\frac{4}{3}$

$\frac{525}{376}$

$\frac{900}{11}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) Shortest wavelength comes from $n_{1} = \infty$ to $n_{2} = 1$ and longest wavelength comes from $n_{1} = 6$ to $n_{2} = 5$ in the given case. Hence

$\frac{1}{λ_{\min}} = R (\frac{1}{1^{2}} - \frac{1}{\infty^{2}}) = R \frac{1}{λ_{\max}} = R (\frac{1}{5^{2}} - \frac{1}{6^{2}}) = R (\frac{36 - 25}{25 \times 36}) = \frac{11}{900} R$

$∴ \frac{λ_{\max}}{λ_{\min}} = \frac{900}{11}$

● ● ● ● ●

In Bohr's model of hydrogen atom, let PE represents potential energy and TE the total energy. In going to a higher level

PE decreases, TE increases

PE increases, TE increases

PE decreases, TE decreases

PE increases, TE decreases

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) As n increases P.E. increases and K.E. decreases.

● ● ● ● ●

According to Bohr's model, the radius of the second orbit of helium atom is

0.53 Å

1.06 Å

2.12 Å

0.265 Å

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) $r = \frac{n^{2}}{Z} (r_{0}); \Rightarrow r_{(n = 2)} = \frac{{(2)}^{2}}{2} \times 0.53 = 1.06$ Å

● ● ● ● ●

The fact that photons carry energy was established by

Doppler's effect

Compton's effect

Bohr's theory

Diffraction of light

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

● ● ● ● ●

An ionic atom equivalent to hydrogen atom has wavelength equal to 1/4 of the wavelengths of hydrogen lines. The ion will be

${He}^{+}$

${Li}^{+ +}$

${Ne}^{9 +}$

${Na}^{10 +}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) $\bar{ν} \propto \frac{1}{λ} \propto Z^{2} \Rightarrow {λZ}^{2} = constant \Rightarrow λ = \frac{λ}{4} Z^{2} \Rightarrow Z = 2$

● ● ● ● ●

The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be

3.4 eV

13.6 eV

54.4 eV

122.4 eV

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$∴ \frac{1}{λ} = {RZ}^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$ gives $Z^{2} = \frac{n_{1}^{2} n_{2}^{2}}{(n_{2}^{2} - n_{1}^{2}) λR}$

On putting values Z = 2
From $E = - \frac{13.6 {(Z)}^{2}}{n^{2}} = \frac{- 13.6 (2)^{2}}{(1)^{2}} = - 54.4 eV$

● ● ● ● ●

An electron in the n = 1 orbit of hydrogen atom is bound by 13.6 eV energy is required to ionize it is

13.6 eV

6.53 eV

5.4 eV

1.51 eV

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) Ionization energy = Binding energy.

● ● ● ● ●

Hydrogen atom emits blue light when it changes from n = 4 energy level to the n = 2 level. Which colour of light would the atom emit when it changes from the n = 5 level to the n = 2 level

Red

Yellow

Green

Violet

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d) In the transition from orbit $5 \to 2$ , more energy is liberated as compared to transition from $4 \to 2$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.