145765
In hydrogen atom spectrum, frequency of $2.7 \times 10^{15} \mathrm{~Hz}$ of electromagnetic wave is emitted when transmission takes place from level 2 to 1. If it moves from 3 to 1 , the frequency emitted will be.
1 $3.2 \times 10^{15} \mathrm{~Hz}$
2 $32 \times 10^{15} \mathrm{~Hz}$
3 $1.6 \times 10^{15} \mathrm{~Hz}$
4 $16 \times 10^{15} \mathrm{~Hz}$
Explanation:
A The frequency $v$ of the emitted by electromagnetic radiation $\mathrm{v}=\operatorname{RcZ}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ When transition takes place $n_{2}=2$ to $n_{1}=1$ $2.7 \times 10^{15}=\operatorname{RcZ}^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $2.7 \times 10^{15}=\operatorname{RcZ}^{2} \times \frac{3}{4}$ When transition takes place from $\mathrm{n}_{2}=3 \rightarrow 0 \mathrm{n}_{1}=1$ $v=\operatorname{RcZ}^{2}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)$ $v=\operatorname{RcZ}^{2}\left(\frac{9-1}{9}\right)$ $v=\operatorname{RcZ}^{2}\left(\frac{8}{9}\right)$ Dividing equation (ii) by equation (i), we get- $\frac{v}{2.7 \times 10^{15}}=\frac{\mathrm{RcZ}^{2} \times \frac{8}{9}}{\mathrm{RcZ}^{2} \times \frac{3}{4}}$ $\frac{v}{2.7 \times 10^{15}}=\frac{8}{9} \times \frac{4}{3}$ $v=\frac{32}{27} \times 2.7 \times 10^{15}$ $=32 \times 0.1 \times 10^{15}$ $v=3.2 \times 10^{15} \mathrm{~Hz}$
UP CPMT-2007
ATOMS
145767
The wavelength of $K_{a}$-line in copper is $1.54 \AA$. The ionization energy of $K$-electron in copper in joule is
145769
Ionisation potential of hydrogen atom is $\mathbf{1 3 . 6}$ $\mathrm{eV}$. Hydrogen atom on the ground state rarely excited by monochromatic radiation of photon $12.1 \mathrm{eV}$. The special line emitted by a hydrogen atom according to Bohr's theory will be
1 One
2 Two
3 Three
4 Four
Explanation:
C Final energy of electron, $-13.6+12.1=-1.5 \mathrm{eV}$ Corresponds for III level i.e $\mathrm{n}=3$ We know that, $\mathrm{E}_{\mathrm{n}}=-\frac{13.6}{\mathrm{n}^{2}}$ $-1.5 \mathrm{eV}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}$ $\mathrm{n}^{2}=\frac{13.6}{1.5}=\frac{136}{15} \square 9$ $\mathrm{n}=3$ i.e. energy of electron in excited state corresponds to third orbit Hence, the possible spectral line are when electron jump from $3^{\text {rd }}$ to $2^{\text {nd }}, 3^{\text {rd }}$ to $1^{\text {st }}, 2^{\text {nd }}$ to $1^{\text {st }}$ thus 3 spectral lines are emitted
JIPMER-2013
ATOMS
145770
Wavelength of light emitted from second orbit to first orbit in a hydrogen atom is
145765
In hydrogen atom spectrum, frequency of $2.7 \times 10^{15} \mathrm{~Hz}$ of electromagnetic wave is emitted when transmission takes place from level 2 to 1. If it moves from 3 to 1 , the frequency emitted will be.
1 $3.2 \times 10^{15} \mathrm{~Hz}$
2 $32 \times 10^{15} \mathrm{~Hz}$
3 $1.6 \times 10^{15} \mathrm{~Hz}$
4 $16 \times 10^{15} \mathrm{~Hz}$
Explanation:
A The frequency $v$ of the emitted by electromagnetic radiation $\mathrm{v}=\operatorname{RcZ}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ When transition takes place $n_{2}=2$ to $n_{1}=1$ $2.7 \times 10^{15}=\operatorname{RcZ}^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $2.7 \times 10^{15}=\operatorname{RcZ}^{2} \times \frac{3}{4}$ When transition takes place from $\mathrm{n}_{2}=3 \rightarrow 0 \mathrm{n}_{1}=1$ $v=\operatorname{RcZ}^{2}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)$ $v=\operatorname{RcZ}^{2}\left(\frac{9-1}{9}\right)$ $v=\operatorname{RcZ}^{2}\left(\frac{8}{9}\right)$ Dividing equation (ii) by equation (i), we get- $\frac{v}{2.7 \times 10^{15}}=\frac{\mathrm{RcZ}^{2} \times \frac{8}{9}}{\mathrm{RcZ}^{2} \times \frac{3}{4}}$ $\frac{v}{2.7 \times 10^{15}}=\frac{8}{9} \times \frac{4}{3}$ $v=\frac{32}{27} \times 2.7 \times 10^{15}$ $=32 \times 0.1 \times 10^{15}$ $v=3.2 \times 10^{15} \mathrm{~Hz}$
UP CPMT-2007
ATOMS
145767
The wavelength of $K_{a}$-line in copper is $1.54 \AA$. The ionization energy of $K$-electron in copper in joule is
145769
Ionisation potential of hydrogen atom is $\mathbf{1 3 . 6}$ $\mathrm{eV}$. Hydrogen atom on the ground state rarely excited by monochromatic radiation of photon $12.1 \mathrm{eV}$. The special line emitted by a hydrogen atom according to Bohr's theory will be
1 One
2 Two
3 Three
4 Four
Explanation:
C Final energy of electron, $-13.6+12.1=-1.5 \mathrm{eV}$ Corresponds for III level i.e $\mathrm{n}=3$ We know that, $\mathrm{E}_{\mathrm{n}}=-\frac{13.6}{\mathrm{n}^{2}}$ $-1.5 \mathrm{eV}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}$ $\mathrm{n}^{2}=\frac{13.6}{1.5}=\frac{136}{15} \square 9$ $\mathrm{n}=3$ i.e. energy of electron in excited state corresponds to third orbit Hence, the possible spectral line are when electron jump from $3^{\text {rd }}$ to $2^{\text {nd }}, 3^{\text {rd }}$ to $1^{\text {st }}, 2^{\text {nd }}$ to $1^{\text {st }}$ thus 3 spectral lines are emitted
JIPMER-2013
ATOMS
145770
Wavelength of light emitted from second orbit to first orbit in a hydrogen atom is
145765
In hydrogen atom spectrum, frequency of $2.7 \times 10^{15} \mathrm{~Hz}$ of electromagnetic wave is emitted when transmission takes place from level 2 to 1. If it moves from 3 to 1 , the frequency emitted will be.
1 $3.2 \times 10^{15} \mathrm{~Hz}$
2 $32 \times 10^{15} \mathrm{~Hz}$
3 $1.6 \times 10^{15} \mathrm{~Hz}$
4 $16 \times 10^{15} \mathrm{~Hz}$
Explanation:
A The frequency $v$ of the emitted by electromagnetic radiation $\mathrm{v}=\operatorname{RcZ}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ When transition takes place $n_{2}=2$ to $n_{1}=1$ $2.7 \times 10^{15}=\operatorname{RcZ}^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $2.7 \times 10^{15}=\operatorname{RcZ}^{2} \times \frac{3}{4}$ When transition takes place from $\mathrm{n}_{2}=3 \rightarrow 0 \mathrm{n}_{1}=1$ $v=\operatorname{RcZ}^{2}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)$ $v=\operatorname{RcZ}^{2}\left(\frac{9-1}{9}\right)$ $v=\operatorname{RcZ}^{2}\left(\frac{8}{9}\right)$ Dividing equation (ii) by equation (i), we get- $\frac{v}{2.7 \times 10^{15}}=\frac{\mathrm{RcZ}^{2} \times \frac{8}{9}}{\mathrm{RcZ}^{2} \times \frac{3}{4}}$ $\frac{v}{2.7 \times 10^{15}}=\frac{8}{9} \times \frac{4}{3}$ $v=\frac{32}{27} \times 2.7 \times 10^{15}$ $=32 \times 0.1 \times 10^{15}$ $v=3.2 \times 10^{15} \mathrm{~Hz}$
UP CPMT-2007
ATOMS
145767
The wavelength of $K_{a}$-line in copper is $1.54 \AA$. The ionization energy of $K$-electron in copper in joule is
145769
Ionisation potential of hydrogen atom is $\mathbf{1 3 . 6}$ $\mathrm{eV}$. Hydrogen atom on the ground state rarely excited by monochromatic radiation of photon $12.1 \mathrm{eV}$. The special line emitted by a hydrogen atom according to Bohr's theory will be
1 One
2 Two
3 Three
4 Four
Explanation:
C Final energy of electron, $-13.6+12.1=-1.5 \mathrm{eV}$ Corresponds for III level i.e $\mathrm{n}=3$ We know that, $\mathrm{E}_{\mathrm{n}}=-\frac{13.6}{\mathrm{n}^{2}}$ $-1.5 \mathrm{eV}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}$ $\mathrm{n}^{2}=\frac{13.6}{1.5}=\frac{136}{15} \square 9$ $\mathrm{n}=3$ i.e. energy of electron in excited state corresponds to third orbit Hence, the possible spectral line are when electron jump from $3^{\text {rd }}$ to $2^{\text {nd }}, 3^{\text {rd }}$ to $1^{\text {st }}, 2^{\text {nd }}$ to $1^{\text {st }}$ thus 3 spectral lines are emitted
JIPMER-2013
ATOMS
145770
Wavelength of light emitted from second orbit to first orbit in a hydrogen atom is
145765
In hydrogen atom spectrum, frequency of $2.7 \times 10^{15} \mathrm{~Hz}$ of electromagnetic wave is emitted when transmission takes place from level 2 to 1. If it moves from 3 to 1 , the frequency emitted will be.
1 $3.2 \times 10^{15} \mathrm{~Hz}$
2 $32 \times 10^{15} \mathrm{~Hz}$
3 $1.6 \times 10^{15} \mathrm{~Hz}$
4 $16 \times 10^{15} \mathrm{~Hz}$
Explanation:
A The frequency $v$ of the emitted by electromagnetic radiation $\mathrm{v}=\operatorname{RcZ}^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)$ When transition takes place $n_{2}=2$ to $n_{1}=1$ $2.7 \times 10^{15}=\operatorname{RcZ}^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)$ $2.7 \times 10^{15}=\operatorname{RcZ}^{2} \times \frac{3}{4}$ When transition takes place from $\mathrm{n}_{2}=3 \rightarrow 0 \mathrm{n}_{1}=1$ $v=\operatorname{RcZ}^{2}\left(\frac{1}{1^{2}}-\frac{1}{3^{2}}\right)$ $v=\operatorname{RcZ}^{2}\left(\frac{9-1}{9}\right)$ $v=\operatorname{RcZ}^{2}\left(\frac{8}{9}\right)$ Dividing equation (ii) by equation (i), we get- $\frac{v}{2.7 \times 10^{15}}=\frac{\mathrm{RcZ}^{2} \times \frac{8}{9}}{\mathrm{RcZ}^{2} \times \frac{3}{4}}$ $\frac{v}{2.7 \times 10^{15}}=\frac{8}{9} \times \frac{4}{3}$ $v=\frac{32}{27} \times 2.7 \times 10^{15}$ $=32 \times 0.1 \times 10^{15}$ $v=3.2 \times 10^{15} \mathrm{~Hz}$
UP CPMT-2007
ATOMS
145767
The wavelength of $K_{a}$-line in copper is $1.54 \AA$. The ionization energy of $K$-electron in copper in joule is
145769
Ionisation potential of hydrogen atom is $\mathbf{1 3 . 6}$ $\mathrm{eV}$. Hydrogen atom on the ground state rarely excited by monochromatic radiation of photon $12.1 \mathrm{eV}$. The special line emitted by a hydrogen atom according to Bohr's theory will be
1 One
2 Two
3 Three
4 Four
Explanation:
C Final energy of electron, $-13.6+12.1=-1.5 \mathrm{eV}$ Corresponds for III level i.e $\mathrm{n}=3$ We know that, $\mathrm{E}_{\mathrm{n}}=-\frac{13.6}{\mathrm{n}^{2}}$ $-1.5 \mathrm{eV}=\frac{-13.6}{\mathrm{n}^{2}} \mathrm{eV}$ $\mathrm{n}^{2}=\frac{13.6}{1.5}=\frac{136}{15} \square 9$ $\mathrm{n}=3$ i.e. energy of electron in excited state corresponds to third orbit Hence, the possible spectral line are when electron jump from $3^{\text {rd }}$ to $2^{\text {nd }}, 3^{\text {rd }}$ to $1^{\text {st }}, 2^{\text {nd }}$ to $1^{\text {st }}$ thus 3 spectral lines are emitted
JIPMER-2013
ATOMS
145770
Wavelength of light emitted from second orbit to first orbit in a hydrogen atom is
