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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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Which of the following spectral series in hydrogen atom gives spectral line of 4860 $\overset{0}{A}$ ?

Lyman

Balmer

Paschen

Brackett

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Explanation

4860 $\overset{0}{A}$ is in visible part of spectrum which is the case for Balmer series.

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Hydrogen $({}_{1}H_{1})$ , Deuterium $({}_{1}H_{2})$ , singly ionised Helium ${({}_{2}H e^{4})}^{+}$ and doubly ionised lithium ${({}_{3}L i^{6})}^{+ +}$ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wave lengths of emitted radiation are $λ_{1}, λ_{2}, λ_{3}$ and $λ_{4}$ respectively then approximately which one of the following is correct [JEE (Main) 2014]

$4 λ_{1} = 2 λ_{2} = 2 λ_{3} = λ_{4}$

$λ_{1} = 2 λ_{2} = 2 λ_{3} = λ_{4}$

$λ_{1} = λ_{2} = 4 λ_{3} = 9 λ_{4}$

$λ_{1} = 2 λ_{2} = 3 λ_{3} = λ_{4}$

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Explanation

$\frac{1}{λ} = R z^{2} (\frac{1}{1^{2}} - \frac{1}{2^{2}}) \frac{1}{λ_{1}} = R {(1)}^{2} (3 / 4), \frac{1}{λ_{2}} = R {(1)}^{2} (3 / 4) \frac{1}{λ_{3}} = R 2^{2} (3 / 4), \frac{1}{λ_{4}} = R 3^{2} (3 / 4) \frac{1}{λ_{1}} = \frac{1}{4 λ_{3}} = \frac{1}{9 λ_{4}} = \frac{1}{λ_{2}}$

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To explain his theory, Bohr used

conservation of linear momentum

conservation of angular momentum

conservation of quantum frequency

conservation of energy

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Explanation

Acoording to Bohr, electron can revolve only in certain discrete non-radiating orbits, called stationary orbits, for which total angular momentum of the revolving electron is an integral multiple of $\frac{h}{2 π}$ , where h is Planck's constant. For orbits conservation of angular momentum is applicable.

$∴$ For any permitted orbit, mvr = $\frac{n h}{2 π}$

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The minimum wavelength of X-rays produced by electrons accelerated by a potential difference of V volt is equal to

$\frac{e V}{h c}$

$\frac{e h}{c V}$

$\frac{h c}{e V}$

$\frac{c V}{e h}$

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Explanation

$E n e r g y E = e V = h v_{m a x} = \frac{h c}{λ_{m i n}} ∴ λ_{m i n} = \frac{h c}{e V}$

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The wavelength of $K_{α}$ X-rays for lead isotopes $P b^{208}, P b^{206}, P b^{204} a r e λ_{1}, λ_{2} a n d λ_{3}$ respectively. Then

$λ_{1} = λ_{2} > λ_{3}$

$λ_{1} > λ_{2} > λ_{3}$

$λ_{1} < λ_{2} < λ_{3}$

$λ_{2} = λ_{1} = λ_{3}$

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Explanation

Since, atomic numbers are same, $λ_{2} = λ_{1} = λ_{3}$

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The longest and shortest wavelength of the Lyman series are (respectively)

$1215 \overset{0}{A}, 500 \overset{0}{A}$

$1215 \overset{0}{A}, 912 \overset{0}{A}$

$912 \overset{0}{A}, 500 \overset{0}{A}$

$912 \overset{0}{A}, 700 \overset{0}{A}$

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Explanation

The Lyman series in the hydrogen atom corresponds to the transitions from higher energy levels to the first energy level (n=1). The longest wavelength in the Lyman series corresponds to the transition from n=2 to n=1, which is 1215 Ã…. The shortest wavelength corresponds to the transition from n=âˆž to n=1, which is 912 Ã….

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$In a hydrogen like atom, when an electron makes transition from third excited state to first excited state, the wavelength of emitted photon is λ_{1}, when the same electron makes a transition from first excited state to ground state, the wavelength of emitted photon is λ_{2}, the value of \frac{λ_{1}}{λ_{2}} is -$

$\frac{1}{2}$

$\frac{1}{4}$

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Explanation

$(2) \frac{1}{λ} = R (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) Case 1 : n_{1} = 2, n_{2} = 4 \frac{1}{λ_{1}} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) \Rightarrow λ_{1} = \frac{16}{3 R} Case 2 : n_{1} = 1, n_{2} = 2 \frac{1}{λ_{2}} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}}) \Rightarrow λ_{2} = \frac{4}{3 R} \frac{λ_{1}}{λ_{2}} = 4$

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The shortest wavelength of Balmer series is about

912 $\overset{o}{A}$

121 $\overset{o}{A}$

3648 $\overset{o}{A}$

4864 $\overset{o}{A}$

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Explanation

$(3) \frac{1}{λ} = R \{\frac{1}{4} - \frac{1}{\infty}\} = 3648 \overset{o}{A}$

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$If λ_{1} be longest wavelength of Lyman series and λ_{2} be the longest wavelength of Balmer series for a hydrogen atom . What is the wavelength of photon emitted when an electron makes a transition from n = 3 to n = 1 in a hydrogen atom ?$

$λ_{1} + λ_{2}$

${(\frac{1}{λ_{1}} + \frac{1}{λ_{2}})}^{- 1}$

$\frac{λ_{1} + λ_{2}}{2}$

$\sqrt{λ_{1} λ_{2}}$

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Explanation

$(2) L y m a n s e r i e s : \frac{1}{λ_{1}} = R [\frac{1}{1^{2}} - \frac{1}{2^{2}}] E n e r g y o f p h o t o n E_{1} = \frac{h c}{λ_{1}} B a l m e r s e r i e s : \frac{1}{λ_{2}} = R [\frac{1}{2^{2}} - \frac{1}{3^{2}}] E n e r g y o f p h o t o n E_{2} = \frac{h c}{λ_{2}} F o r t r a n s i t i o n f r o m n = 3 t o n = 1, E = E_{3 \to 2} + E_{2 \to 1} = E_{2} + E_{1} \frac{hc}{λ_{3}} = \frac{hc}{λ_{1}} + \frac{hc}{λ_{2}}$

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Yes. You can attempt every Physics 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.