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When hydrogen atom is in its first excited level, its radius is [1997]

four times, its ground state radius

twice, its ground state radius

same as its ground state radius

half of its ground state radius

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Explanation

The radius of nth Bohr's orbit of hydrogen and hydrogen like atom

$r_{n} = \frac{ε_{o} n^{2} h^{2}}{π m e^{2} Z} ∴ r_{n} = \frac{n^{2} a_{0}}{Z} o r r_{n} α \frac{n^{2}}{Z}$

For ground state, n = 1

Atomic number, Z = 1

For first excited state, n = 2

$∴ \frac{r_{2}}{r_{1}} = {(\frac{2}{1})}^{2} = 4 o r r_{2} = 4 r_{1}$

Therefore, radius of first excited state is 4 times than that of ground state radius in H-atom.

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The ionisation potential of helium atom is 24.6 volt, the energy required to ionise it will be

24.6 eV

24.6 volt

13.6 volt

13..6 eV

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Explanation

Energy, W = qV, ionisation energy in eV = Ionisation Potential in Volt

$∴$ 24.6 eV energy is required to ionise helium atom.

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Ratio of the wavelength of first line of Lyman series and first line of Balmer series is [EAMCET (Engg. 1995; MP PMT 1997; UPSEAT 2004; KCET 2006; NEET 2013]

1 : 3

27 : 5

5 : 27

4 : 9

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Explanation

$\frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

For first line of Lyman series $n_{1}$ = 1 and $n_{2}$ = 3

For first line of Balmer series $n_{1}$ = 2 and $n_{2}$ = 3

So, $\frac{λ_{L y m a n}}{λ_{B a l m e r}} = \frac{5}{27}$

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The ionization energy of 10 times ionized sodium atom is:

13.6 eV

$13.6 \times 11 e V$

$\frac{13.6}{11} e V$

$13.6 \times {(11)}^{2} e V$

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Explanation

$(E_{i o n}) N a = Z^{2} (E_{i o n}) H = {(11)}^{2} \times 13.6 e V$

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The ionisation energy of hydrogen atom is 13.6 eV, the ionisation energy of helium atom would be [1988]

13.6 eV

27.2 eV

6.8 eV

54.4 eV

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Explanation

Ionisation energy is defined as the energy required to knock an electron completely out of an isolated gaseous atom. When electron is raised to the orbit n = $\infty$ , it will be completely out of the atom. Ionisation energy of helium,

$E = \frac{2 π^{2} {mZ}^{2} k^{2} e^{4}}{h^{2}} [\frac{1}{1^{2}} - \frac{1}{\infty}] = \frac{2 π^{2} m {(2)}^{2} k^{2} e^{4}}{h^{2}} [\frac{1}{1^{2}} - \frac{1}{\infty}] I o n i s a t i o n e n e r g y f o r h y d r o g e n a t o m = 13.6 e V = \frac{2 π^{2} {mk}^{2} e^{2}}{h^{2}} \times 4 \times [\frac{1}{1^{2}} - \frac{1}{\infty^{2}}] \Rightarrow 4 \times 13.6 = 54.4 e V$

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The photon radiated from hydrogen corresponding to the line of Lyman series is absorbed by a hydrogen-like atom X in the excited state. As a result the hydrogen-like atom X makes a transition to $n^{t h}$ orbit. Then:

X = $H e^{+}, n = 4$

X = $L i^{+ +}$ , n = 6

X = $H e^{+}, n = 9$

X = $L i^{+ +}$ , n = 9

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Explanation

Energy of $n^{t h}$ state in hydrogen is same as energy of $3 n^{t h}$ state in $L i^{+ +}$ .

$∴$ 3 $\to$ 1 transition in H would give same energy as the

3 $\times$ 3 $\to$ 1 $\times$ 3 transition in $L i^{+ +}$

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If an electron in an hydrogen atom jumps from an orbit $n_{i} = 3$ to an orbit with level $n_{f} = 2$ , the frequency of the emitted radiation is

$v = \frac{36 C}{5 R}$

v = $\frac{C R}{6}$

v = $\frac{5 R C}{36}$

v = $\frac{6 C}{R}$

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Explanation

$\frac{1}{λ} = R (\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}) C = ν λ; \frac{1}{λ} = \frac{ν}{C} \frac{ν}{C} = R (\frac{1}{n_{i}^{2}} - \frac{1}{n_{f}^{2}}) ν = C R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5}{36} C R$

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The energy of a hydrogen atom in its ground state is -13.6 eV. The energy of the level corresponding to the quantum number n = 5 is

- 0.54 eV

- 5.40 eV

- 0.85 eV

-2.72 eV

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Explanation

$E_{n} = - \frac{13.6}{n^{2}} e V = \frac{13.6}{5^{2}} e V = - 0.54 e V$

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The wave number of a photon in Brackett series of hydrogen atom is $\frac{9}{400} R$ . The electron has transited from the orbit havit quantum number [Odisha JEE 2012]

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Explanation

$F o r B r a c k e t t s e r i e s, \bar{v} = R (\frac{1}{4^{2}} - \frac{1}{n^{2}}) \frac{9}{400} R = R (\frac{1}{16} - \frac{1}{n^{2}}) o r \frac{9}{400} = \frac{1}{16} - \frac{1}{n^{2}} \frac{1}{n^{2}} = \frac{1}{16} - \frac{9}{400} = \frac{16}{400} n^{2} = \frac{400}{16} o r n = \frac{20}{4} = 5$

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In an experiment to determine the e/m value for an electron using Thomson's method the electrostatic deflection plates were 0.01 m apart and had a potential difference of 200 volts applied. Then the electric field strength between the plates is

$1 \times 10^{4} V / m$

$2 \times 10^{4} V / m$

$4 \times 10^{4} V / m$

$5 \times 10^{5} V / m$

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Explanation

$E = \frac{V}{d} = \frac{200}{0.01} = 20, 000 V / m = 2 \times 10^{4} V / m$

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