13. Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr's atomic model? ⇒ (AIPMT 2011 Mains)

A. 0.65 eV

B. 1.9 eV

C. 11.1 eV

D. 13.6 eV

The Correct Answer is Option (C)

E_n = $- \frac{13.6}{n^{2}}$

$∴$ E₁ = –13.6 eV

E₂ = –3.4 eV

E₃ = –1.5 eV

E₄ = –0.85 eV

$∴$ E₃ – E₂ = –1.5 – (–3.4) = 1.9 eV

E₄ – E₃ = –0.85 – (–1.5) = 0.65 eV

Obviously, difference of 11.1 eV is not possible.

14. The energy of a hydrogen atom in the ground state is $-$ 13.6 eV. The energy of a He⁺ ion in the first excited state will be⇒ (AIPMT 2010 Prelims)

A. $-$ 13.6 eV

B. $-$ 27.2 eV

C. $-$ 54.4 eV

D. $-$ 6.8 eV

The Correct Answer is Option (A)

Energy of a H-like atom in it's n^th state is given by,

E_n = $- \frac{13.6}{n^{2}} Z^{2}$

For, first excited state of He⁺, n = 2, Z = 2

$∴$ E_He⁺ = $- \frac{13.6}{2^{2}} \times 2^{2}$ = -13.6 eV

15. The energy of a hydrogen atom in the ground state is $-$ 13.6 eV. The energy of a He⁺ ion in the first excited state will be ⇒ (AIPMT 2010 Prelims)

A. $-$ 13.6 eV

B. $-$ 27.2 eV

C. $-$ 54.4 eV

D. $-$ 6.8 eV

The Correct Answer is Option (A)

Currently no explanation available

16. The ground state energy of hydrogen atom is $-$ 13.6 eV. When its electron is in the first excited state, its excitation energy is ⇒ (AIPMT 2008)

A. 10.2 eV

B. 0

C. 3.4 eV

D. 6.8 eV

The Correct Answer is Option (A)

E_n = $- \frac{13.6}{n^{2}}$

E₁ = -13.6 eV

E₂ = $- \frac{13.6}{2^{2}}$ = -3.4 eV

$∴$ $Δ$ E = ( – 3.4) – ( – 13.6) = 10.2 eV.

17. In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential $V$ and then made to describe semicircular paths of radius R using a magnetic field B. If V and B are kept constant, the ratio $(\frac{c h \arg e o n t h e i o n}{m a s s o f t h e i o n})$ will be proportional to ⇒ (AIPMT 2007 )

A. 1/R²

B. R²

C. R

D. 1/R

The Correct Answer is Option (A)

In mass spectrometer, when ions are accelerated through potential V,

$\frac{1}{2} m v^{2} = q V$ ....(1)

As the magnetic field curves the path of the ions in a semicircular orbit.

$∴$ Bqv = $\frac{m v^{2}}{R}$

$\Rightarrow$ v = $\frac{B q R}{m}$ .....(2)

Now by substituting (ii) in (i),

$\frac{1}{2} m {(\frac{B q R}{m})}^{2} = q$

$\Rightarrow$ $\frac{q}{m} = \frac{2 V}{B^{2} R^{2}}$

As V and B are constants,

$∴$ $\frac{q}{m}$ $\propto$ $\frac{1}{R^{2}}$

18. The total energy of electron in the ground state of hydrogen atom is $-$ 13.6 eV. The kinetic energy of an electron in the first excited state is ⇒ (AIPMT 2007)

A. 6.8 eV

B. 13.6 eV

C. 1.7 eV

D. 3.4 eV.

The Correct Answer is Option (D)

E_n = $- \frac{13.6}{n^{2}}$

E₁ = -13.6 eV

E₂ = $- \frac{13.6}{2^{2}}$ = -3.4 eV

Kinetic energy of an electron in the first excited state is

K = –E₂ = 3.4 eV.

19. The total energy of an electron in the first excited state of hydrogen atom is about $-$ 3.4 eV. Its kinetic energy in this state is ⇒ ( AIPMT 2005)

A. 3.4 eV

B. 6.8 eV

C. $-$ 3.4 eV

D. $-$ 6.8 eV

The Correct Answer is Option (A)

Apply Bohr's Atomic model for H-atom

K.E = -T.E = +3.4 eV

20. The Bohr model of atoms ⇒ (AIPMT 2004)

A. Assumes that the angular momentum of electrons is quantized.

B. Uses Einstein's photoelectric equation.

C. Predicts continuous emission spectra fror atoms.

D. Predicts the same emission spectra for all types of atoms.

The Correct Answer is Option (A)

Bohr model of atoms assumes that the angular momentum of electrons is quantised.

21. An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force $\vec{F}$ between the two is ⇒ ( AIPMT 2003)

A. $K \frac{e^{2}}{r^{2}} \hat{r}$

B. $- K \frac{e^{2}}{r^{3}} \hat{r}$

C. $K \frac{e^{2}}{r^{3}} \hat{r}$

D. $- K \frac{e^{2}}{r^{2}} \hat{r}$

The Correct Answer is Option (D)

The charge on hydrogen nucleus q₁ = +e

charge on electron, q₂ = –e

Coulomb force, F = $K \frac{(+ e) (- e)}{r^{2}}$

= $- \frac{K e^{2}}{r^{3}} \vec{r}$

= $- \frac{K e^{2}}{r^{2}} \hat{r}$

22. In which of the following systems will the radius of the first orbit (n = 1) be minimum ? ⇒ ( AIPMT 2003)

A. doubly ionized lithium

B. singly ionized helium

C. deuterium atom

D. hydrogen atom

The Correct Answer is Option (A)

As Radius of first orbit, r $\propto$ $\frac{1}{Z}$

for doubly ionized lithium Z (= 3) will be maximum, hence for doubly ionized lithium, r will be minimum.

23. The energy of hydrogen atom in n^th orbit is E_n then the energy in n^th orbit of singly ionised helium atom will be ⇒ (AIPMT 2001)

A. 4E_n

B. E_n/4

C. 2E_n

D. E_n/2

The Correct Answer is Option ()

As E_n = $\frac{- 2 π^{2} m K^{2} Z^{2} e^{4}}{n^{2} h^{2}}$

For helium Z = 2. Hence requisite answer is 4E_n.

24. Maximum frequency of emission is obtained for the transition⇒ (AIPMT 2000)

A. n = 2 to n = 1

B. n = 6 to n = 2

C. n = 1 to n = 2

D. n = 2 to n = 6.

The Correct Answer is Option (A)

$ν$ $\propto$ $[\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

Maximum frequency of emission happens when n₁ = 1 and n₂ = 2

⇒ Atom

Topic 2 : Bohr's Atomic Model