MCQ 511 Mark
Every series of hydrogen spectrum has an upper and lower limit in wavelength. The spectral series which has an upper limit of wavelength equal to $18752 \mathring A$ isv
Answer$ \frac{1}{\lambda}=R\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \Rightarrow \frac{1}{n_1^2}-\frac{1}{n_2^2}=\frac{1}{R \lambda}$
$ =\frac{1}{1.097 \times 10^7 \times 18752 \times 10^{-10}}=0.0486=\frac{7}{144} . \text { But } $
$ \frac{1}{3^2}-\frac{1}{4^2}=\frac{7}{144} \Rightarrow n_1=3 \text { and } n=4 \quad \text { (Paschen series) }$
View full question & answer→MCQ 521 Mark
The energy required to remove an electron in a hydrogen atom from $n=10$ state is
- A
$13.6\ eV$
- B
$1.36\ eV$
- ✓
$0.136\ eV$
- D
$0.0136\ eV$
AnswerCorrect option: C. $0.136\ eV$
Energy required $=\frac{13.6}{n^2}=\frac{13.6}{10^2}=0.136\ \mathrm{eV}$
View full question & answer→MCQ 531 Mark
In the $n$ orbit, the energy of an electron $E_n=-\frac{13.6}{n^2}\ eV$ for hydrogen atom. The energy required to take the electron from first orbit to second orbit will be
- ✓
$10.2 eV$
- B
$12.1 eV$
- C
$13.6 eV$
- D
$3.4 eV$
AnswerCorrect option: A. $10.2 eV$
View full question & answer→MCQ 541 Mark
The kinetic energy of the electron in an orbit of radius $r$ in hydrogen atom is $(e=$ electronic charge $)$ (Rydberg constant $R=1.097 \times 10^7$ per metre)
- A
$\frac{e^2}{r^2}$
- ✓
$\frac{e^2}{2 r}$
- C
$\frac{e^2}{r}$
- D
$\frac{e^2}{2 r^2}$
AnswerCorrect option: B. $\frac{e^2}{2 r}$
Potential energy of electron in $r$ orbit of radius $r$ in $H$-atom
$U=-\frac{e^2}{r} \text { (in CGS) } $
$ \because \text { K.E. }=\frac{1}{2} \mid \text { P.E. } \mid \Rightarrow K=\frac{e^2}{2 r}$
View full question & answer→MCQ 551 Mark
The wavelength of yellow line of sodium is $5896 \mathring A$. Its wave number will be
AnswerCorrect option: B. $16961$ per $cm$
Wave number $\bar{v}=\frac{1}{\lambda}=\frac{1}{5896 \times 10^{-8}}=16961$ per $\mathrm{cm}$
View full question & answer→MCQ 561 Mark
As the electron in Bohr orbit of Hydrogen atom passes from state $n=2$ to $n=1$, the kinetic energy $K$ and potential energy $U$ change as
- A
$K$ two$-$fold, $U$ four$-$fold
- B
$K$ four$-$fold, $U$ two$-$fold
- ✓
$K$ four$-$fold, $U$ also four$-$fold
- D
$K$ two$-$fold, $U$ also two$-$fold
AnswerCorrect option: C. $K$ four$-$fold, $U$ also four$-$fold
$K$ four$-$fold, $U$ also four$-$fold
View full question & answer→MCQ 571 Mark
The time of revolution of an electron around a nucleus of charge $Z e$ in $\pi$ Bohr orbit is directly proportional to
- A
$n$
- ✓
$\frac{n^3}{Z^2}$
- C
$\frac{n^2}{Z}$
- D
$\frac{Z}{n}$
AnswerCorrect option: B. $\frac{n^3}{Z^2}$
$ T=\frac{2 \pi r}{v} ; r=\text { radius of } \pi \text { orbit }=\frac{n^2 h^2}{\pi m Z e^2} $
$ v=\text { speed of } e^{-} \text {in } r \text { orbit }=\frac{z e^2}{2 \varepsilon_0 n h} $
$ \therefore T=\frac{4 \varepsilon_0^2 n^3 h^3}{m Z^2 e^4} \Rightarrow T \propto \frac{n^3}{Z^2}$
View full question & answer→MCQ 581 Mark
The radius of the first $($lowest$)$ orbit of the hydrogen atom is $a_0$. The radius of the second $($next higher$)$ orbit will be
- ✓
$4 a_0$
- B
$6 a_0$
- C
$8 a_0$
- D
$10 a_0$
AnswerCorrect option: A. $4 a_0$
$4 a_0$
View full question & answer→MCQ 591 Mark
For principal quantum number $n=3$, the possible values of orbital quantum number 'I are
- A
$1, 2, 3$
- B
$0,1,2,3$
- ✓
$0,1,2$
- D
$-1,0,+1$
AnswerCorrect option: C. $0,1,2$
For $M$ shell $(n=3)$, orbital quantum number $I=0,1,2$.
View full question & answer→MCQ 601 Mark
The order of the size of nucleus and Bohr radius of an atom respectively are
- ✓
$10^{-14} m , 10^{-10} m$
- B
$10^{-10} m , 10^{-8} m$
- C
$10^{-20} m , 10^{-16} m$
- D
$10^{-8} m , 10^{-6} m$
AnswerCorrect option: A. $10^{-14} m , 10^{-10} m$
$10^{-14} m , 10^{-10} m$
View full question & answer→MCQ 611 Mark
The ratio of minimum to maximum wavelength in Balmer series is
- A
$5: 9$
- ✓
$5: 36$
- C
$1: 4$
- D
$3: 4$
AnswerCorrect option: B. $5: 36$
$5: 36$
View full question & answer→MCQ 621 Mark
According to Bohr's theory the moment of momentum of an electron revolving in second orbit of hydrogen atom will be
- A
$2 \pi h$
- B
$\pi h$
- ✓
$\frac{h}{\pi}$
- D
$\frac{2 h}{\pi}$
AnswerCorrect option: C. $\frac{h}{\pi}$
Angular momentum $L=n\left(\frac{h}{2 \pi}\right)$
For this case $n=2$, hence $L=2 \times \frac{h}{2 \pi}=\frac{h}{\pi}$
View full question & answer→MCQ 631 Mark
The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is
- A
$\frac{4}{3}$
- B
$\frac{525}{376}$
- C
$25$
- ✓
$\frac{900}{11}$
AnswerCorrect option: D. $\frac{900}{11}$
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 $\frac{1}{\lambda_{\min }}=R\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)=R$
$ \frac{1}{\lambda_{\max }}=R\left(\frac{1}{5^2}-\frac{1}{6^2}\right)=R\left(\frac{36-25}{25 \times 36}\right)=\frac{11}{900} R$
$ \therefore \frac{\lambda_{\max }}{\lambda_{\min }}=\frac{900}{11}$
View full question & answer→MCQ 641 Mark
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
- A
$v$
- B
$\frac{22}{5} v$
- C
$\frac{5}{2} v$
- ✓
$\frac{2}{5} v$
AnswerCorrect option: D. $\frac{2}{5} v$
$v \propto \frac{1}{n} \Rightarrow \frac{v_5}{v_2}=\frac{2}{5} \Rightarrow v_5=\frac{2}{5} v_2=\frac{2}{5} v$
View full question & answer→MCQ 651 Mark
The absorption transitions between the first and the fourth energy states of hydrogen atom are $3.$ The emission transitions between these states will be
AnswerBy using $N_E=\frac{n(n-1)}{2} \Rightarrow N_E=\frac{4(4-1)}{2}=6$
View full question & answer→MCQ 661 Mark
In the lowest energy level of hydrogen atom, the electron has the angular momentum
- A
$\pi / h$
- B
$h / \pi$
- ✓
$h / 2 \pi$
- D
$2 \pi / h$
AnswerCorrect option: C. $h / 2 \pi$
$m v r=\frac{n h}{2 \pi}$, for $n=1$ it is $\frac{h}{2 \pi}$
View full question & answer→MCQ 671 Mark
In a hydrogen atom, which of the following electronic transitions would involve the maximum energy change
- A
From $n=2$ to $n=1$
- ✓
From $n=3$ to $n=1$
- C
From $n=4$ to $n=2$
- D
From $n=3$ to $n=2$
AnswerCorrect option: B. From $n=3$ to $n=1$
From $n=3$ to $n=1$
View full question & answer→MCQ 681 Mark
Which of the following transitions in a hydrogen atom emits photon of the highest frequency
- ✓
$n=1$ to $n=2$
- B
$n=2$ to $n=1$
- C
$n=2$ to $n=6$
- D
$n=6$ to $n=2$
AnswerCorrect option: A. $n=1$ to $n=2$
View full question & answer→MCQ 691 Mark
In any Bohr orbit of the hydrogen atom, the ratio of kinetic energy to potential energy of the electron is
- A
$1 / 2$
- B
$2$
- ✓
$-1 / 2$
- D
$-2$
AnswerCorrect option: C. $-1 / 2$
$K . E=\frac{k Z e^2}{2 r}$ and P.E. $=-\frac{k Z e^2}{r} ; \therefore \frac{K . E .}{P . E .}=-\frac{1}{2}$.
View full question & answer→MCQ 701 Mark
Which of the following is true
- A
Lyman series is a continuous spectrum
- ✓
Paschen series is a line spectrum in the infrared
- C
Balmer series is a line spectrum in the ultraviolet
- D
The spectral series formula can be derived from the Rutherford model of the hydrogen atom
AnswerCorrect option: B. Paschen series is a line spectrum in the infrared
Paschen series lies in the infrared region.
View full question & answer→MCQ 711 Mark
The Lyman series of hydrogen spectrum lies in the region
AnswerLyman series lies in the UV region.
View full question & answer→MCQ 721 Mark
In the Bohr's hydrogen atom model, the radius of the stationary orbit is directly proportional to ( $n=$ principle quantum number)
- A
$n^{-1}$
- B
$n$
- C
$n^{-2}$
- ✓
$n^2$
AnswerBohr radius $r=\frac{\varepsilon_0 n^2 h^2}{\pi Z m e^2} ;\ \therefore r \propto n^2$
View full question & answer→MCQ 731 Mark
The first line in the lyman series has wavelength $\lambda$. The wavelength of the first line in Balmer series is
- A
$\frac{2}{9} \lambda$
- B
$\frac{9}{2} \lambda$
- C
$\frac{5}{27} \lambda$
- ✓
$\frac{27}{5} \lambda$
AnswerCorrect option: D. $\frac{27}{5} \lambda$
For first line in Lyman series $\lambda_{L_1}=\frac{4}{3 R}$
For first line in Balmer series $\lambda_{B_1}=\frac{36}{5 R}$
From equation (i) and (ii)
$\frac{\lambda_{B_1}}{\lambda_{L_1}}=\frac{27}{5} \Rightarrow \lambda_{B_1}=\frac{27}{5} \lambda_{L_1} \Rightarrow \lambda_{B_1}=\frac{27}{5} \lambda$
View full question & answer→MCQ 741 Mark
What is the ratio of wavelength of radiations emitted when an electron in hydrogen atom jump from fourth orbit to second orbit and from third orbit to second orbit
- A
$27: 25$
- ✓
$20: 27$
- C
$20: 25$
- D
$25: 27$
AnswerCorrect option: B. $20: 27$
$20: 27$
View full question & answer→MCQ 751 Mark
Energy of the electron in $n$ orbit of hydrogen atom is given by $E_n=-\frac{13.6}{n^2} eV$. The amount of energy needed to transfer electron from first orbit to third orbit is
- A
$13.6 eV$
- B
$3.4 eV$
- ✓
$12.09 eV$
- D
$1.51 eV$
AnswerCorrect option: C. $12.09 eV$
For $n=1, E_1=-\frac{13.6}{(1)^2}=-13.6 \mathrm{eV}$ and for $n=3, E_3=-\frac{13.6}{(3)^2}=-1.51 \mathrm{eV}$
So required energy$=E_3-E_1=-1.51-(-13.6)=12.09 \mathrm{eV}$
View full question & answer→MCQ 761 Mark
The ratio of speed of an electron in ground state in Bohrs first orbit of hydrogen atom to velocity of light in air is
- A
$\frac{e^2}{2 \varepsilon_0 h c}$
- ✓
$\frac{2 e^2 \varepsilon_0}{h c}$
- C
$\frac{e^3}{2 \varepsilon_0 h c}$
- D
$\frac{2 \varepsilon_0 h c}{e^2}$
AnswerCorrect option: B. $\frac{2 e^2 \varepsilon_0}{h c}$
$\frac{2 e^2 \varepsilon_0}{h c}$
View full question & answer→MCQ 771 Mark
In Bohr model of the hydrogen atom, the lowest orbit corresponds to
MCQ 781 Mark
In which of the following systems will the radius of the first orbit $(n=1)$ be minimum
Answer$r \propto \frac{1}{Z}$, for double ionized lithium $Z(=3)$ will be maximum. So $r$ will be minimum
View full question & answer→MCQ 791 Mark
Radius of the first orbit of the electron in a hydrogen atom is $0.53 \mathring A$. So, the radius of the third orbit will be
AnswerCorrect option: C. $4.77 \mathring A$
$r_n \propto n^2 \Rightarrow \frac{r_3}{r_1}=\frac{3^2}{1} \Rightarrow r_3=9 r_1=9 \times 0.53=4.77 \mathring A$
View full question & answer→MCQ 801 Mark
According to the Rutherford's atomic model, the electrons inside the atom are
MCQ 811 Mark
The energy of hydrogen atom in its ground state is $-13.6\ eV$. The energy of the level corresponding to the quantum number $n$ is equal $5$ is
- A
$-5.40\ eV$
- B
$-2.72\ eV$
- C
$-0.85\ eV$
- ✓
$-0.54\ eV$
AnswerCorrect option: D. $-0.54\ eV$
$E_n=-\frac{13.6}{n^2} \mathrm{eV} \Rightarrow E_5=\frac{-13.6}{5^2}=\frac{-13.6}{25}=-0.54\ \mathrm{eV}$
View full question & answer→MCQ 821 Mark
Bohr's atom model assumes
- A
The nucleus is of infinite mass and is at rest
- B
Electrons in a quantized orbit will not radiate energy
- C
Mass of electron remains constant
- ✓
View full question & answer→MCQ 831 Mark
Which state of triply ionised Baryllium $\left(B e^{+++}\right)$has the same orbital radius as that of the ground state of hydrogen
AnswerRadius of $\pi$ orbit for any hydrogen like atom $r_n=r_0\left(\frac{n^2}{Z}\right)\left(r_0=\right.$ radius of first orbit of $\mathrm{H}_2$-atom $)$
If $r_n=r_0 \Rightarrow n=\sqrt{Z}$.
For $B e, Z=4 \Rightarrow n=2$.
View full question & answer→MCQ 841 Mark
The colour of the second line of Balmer series is
MCQ 851 Mark
The de-Broglie wavelength of an electron in the first Bohr orbit is
- A
Equal to one fourth the circumference of the first orbit
- B
Equal to half the circumference of the first orbit
- C
Equal to twice the circumference of the first orbit
- ✓
Equal to the circumference of the first orbit
AnswerCorrect option: D. Equal to the circumference of the first orbit
$ m v r_n=\frac{n h}{2 \pi} \Rightarrow p r_n=\frac{n h}{2 \pi} \Rightarrow \frac{h}{\lambda} \times r_n=\frac{n h}{2 \pi} $
$ \Rightarrow \lambda=\frac{2 \pi r_n}{n}, \text { for first orbit } n=1 \text { so } \lambda=2 \pi r_1 $
$ =\text { circumference of first orbit }$
View full question & answer→MCQ 861 Mark
Whenever a hydrogen atom emits a photon in the Balmer series
- A
It need not emit any more photon
- B
It may emit another photon in the Paschen series
- ✓
It must emit another photon in the Lyman series
- D
It may emit another photon in the Balmer series
AnswerCorrect option: C. It must emit another photon in the Lyman series
Since in spectral series of hydrogen atom, Lymen series lies lower Balmer series.
View full question & answer→MCQ 871 Mark
If $R$ is the Rydberg's constant for hydrogen the wave number of the first line in the lyman series will be
- A
$\frac{R}{4}$
- ✓
$\frac{3 R}{4}$
- C
$\frac{R}{2}$
- D
$2 R$
AnswerCorrect option: B. $\frac{3 R}{4}$
For Lyman series$\bar{v}=\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$ here $n=2,3,4,5 \ldots ..$
For first line$\bar{v}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \Rightarrow \bar{v}=R\left(1-\frac{1}{4}\right)=\frac{3 R}{4}$
View full question & answer→MCQ 881 Mark
In Bohr's model of hydrogen atom, let $P E$ represents potential energy and $T E$ the total energy. In going to a higher level
- A
$P E$ decreases, $T E$ increases
- ✓
$P E$ increases, $T E$ increases
- C
$P E$ decreases, $T E$ decreases
- D
$P E$ increases, $T E$ decreases
AnswerCorrect option: B. $P E$ increases, $T E$ increases
As $n$ increases P.E. increases and K.E. decreases.
View full question & answer→MCQ 891 Mark
Which one of these is non-divisible
MCQ 901 Mark
The ratio of the frequencies of the long wavelength limits of Lyman and Balmer series of hydrogen spectrum is
- ✓
$27: 5$
- B
$5: 27$
- C
$4: 1$
- D
$1: 4$
AnswerCorrect option: A. $27: 5$
For Lyman series$v_{\text {Lymen }}=\frac{c}{\lambda_{\max }}=R c\left[\frac{1}{(1)^2}-\frac{1}{(2)^2}\right]=\frac{3 R C}{4}$
For Balmer series $ v_{\text {Balmer }}=\frac{c}{\lambda_{\max }}=R c\left[\frac{1}{(2)^2}-\frac{1}{(3)^2}\right]=\frac{5 R C}{36} $
$ \therefore \frac{v_{\text {Lymen }}}{v_{\text {Balmer }}}=\frac{27}{5}$
View full question & answer→MCQ 911 Mark
Which of the following phenomena suggests the presence of electron energy levels in atoms
MCQ 921 Mark
In the following atoms and moleculates for the transition from $n=2$ to $n=1$, the spectral line of minimum wavelength will be produced by
Answer$\frac{1}{\lambda}=R Z^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$For di-ionised lithium the value of $Z$ is maximum.
View full question & answer→MCQ 931 Mark
The fact that photons carry energy was established by
MCQ 941 Mark
The electron in a hydrogen atom makes a transition $n_1 \rightarrow n_2$, where $n_1$ and $n_2$ are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the initial state is eight times that in the final state. The possible values of $n$ and $n$ are
- ✓
$n_1=4, n_2=2$
- B
$n_1=8, n_2=2$
- C
$n_1=8, n_2=1$
- D
$n_1=6, n_2=3$
AnswerCorrect option: A. $n_1=4, n_2=2$
$n_1=4, n_2=2$
View full question & answer→MCQ 951 Mark
As per Bohr model, the minimum energy (in $eV$ ) required to remove an electron from the ground state of doubly ionized $L i$ atom $(Z=3)$ is
- A
$1.51$
- B
$13.6$
- C
$40.8$
- ✓
$122.4$
AnswerCorrect option: D. $122.4$
$E=-Z^2 \times 13.6 \mathrm{eV}=-9 \times 13.6 \mathrm{eV}=-122.4 \mathrm{eV}$
So ionisation energy $=+122.4 \mathrm{eV}$.
View full question & answer→MCQ 961 Mark
If in nature there may not be an element for which the principal quantum number $n>4$, then the total possible number of elements will be
AnswerFor $n=1$, maximum number of states $=2 n^2=2$ and for $n=2$, 3,4 , maximum number of states would be $8,18,32$ respectively, Hence number of possible elements$=2+8+18+32=60 .$
View full question & answer→MCQ 971 Mark
The ratio of the largest to shortest wavelengths in Lyman series of hydrogen spectra is
- A
$\frac{25}{9}$
- B
$\frac{17}{6}$
- C
$\frac{9}{5}$
- ✓
$\frac{4}{3}$
AnswerCorrect option: D. $\frac{4}{3}$
For Lyman series $\frac{1}{\lambda_{\max }}=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3}{4} R$ and$\frac{1}{r_{\min }}=R\left\lceil\frac{1}{1^2}-\frac{1}{2}\right\rceil=\frac{R}{1} \Rightarrow \frac{\lambda_{\max }}{r_{\min }}=\frac{4}{3}$
View full question & answer→MCQ 981 Mark
The first member of the Paschen series in hydrogen spectrum is of wavelength $18,800 \mathring A$. The short wavelengths limit of Paschen series is
- A
$1215 \mathring A$
- B
$6560 \mathring A$
- ✓
$8225 \mathring A$
- D
$12850 \mathring A$
AnswerCorrect option: C. $8225 \mathring A$
For Paschen series $\bar{v}=\frac{1}{\lambda}=R\left[\frac{1}{3^2}-\frac{1}{n^2}\right] ; n=4,5,6 \ldots$.
For first member of Paschen series $n=4$
$ \frac{1}{\lambda_1}=R\left[\frac{1}{3^2}-\frac{1}{4^2}\right] \Rightarrow \frac{1}{\lambda_1}=\frac{7 R}{144} $
$ \Rightarrow R=\frac{144}{7 \lambda_1}=\frac{144}{7 \times 18800 \times 10^{-10}}=1.1 \times 10^{-7}$
For shortest wave length $n=\infty$ So $\frac{1}{\lambda}=R\left[\frac{1}{3^2}-\frac{1}{\infty^2}\right]=\frac{R}{9}$
$\Rightarrow \lambda=\frac{9}{R}=\frac{9}{1.1 \times 10^{-7}}=8.225 \times 10^{-7} \mathrm{~m}=8225 \mathring A$
View full question & answer→MCQ 991 Mark
In hydrogen atom, if the difference in the energy of the electron in $n=2$ and $n=3$ orbits is $E$, the ionization energy of hydrogen atom is
- A
$13.2\ E$
- ✓
$7.2\ E$
- C
$5.6\ E$
- D
$3.2\ E$
AnswerCorrect option: B. $7.2\ E$
$\text { Energy } E=K\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right] \quad(K=\text { constant }) $
$ n=2 \text { and } n=3 \text {, so } E=K\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=K\left[\frac{5}{36}\right]$
For removing an electron $n=1$ to $n_2=\infty$Energy $E_1=K[1]=\frac{36}{5} E=7.2 E$
$\therefore \text { lonization energy }=7.2 E$
View full question & answer→MCQ 1001 Mark
The ratio of the longest to shortest wavelengths in Brackett series of hydrogen spectra is
- ✓
$\frac{25}{9}$
- B
$\frac{17}{6}$
- C
$\frac{9}{5}$
- D
$\frac{4}{3}$
AnswerCorrect option: A. $\frac{25}{9}$
$\frac{25}{9}$
View full question & answer→