145520
The shortest wavelength in the Balmer series of hydrogen atom spectrum is [For hydrogen, Rydberg constant = $1.097 \times 10^{7} \mathrm{~m}^{-1}$ ]
1 $91.2 \mathrm{~nm}$
2 $364.6 \mathrm{~nm}$
3 $820.4 \mathrm{~nm}$
4 $2278.9 \mathrm{~nm}$
Explanation:
B We know that, $\mathrm{R}=1.097 \times 10^{7} \mathrm{~m}^{-1}$ The shortest wavelength of Balmer series $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda}=\frac{1.097 \times 10^{7}}{4}$ $\lambda=\frac{4}{1.097} \times 10^{-7} \mathrm{~m}$ $\lambda=3.6463 \times 10^{-7}=364.6 \times 10^{-9} \mathrm{~m}$ $\lambda=364.6 \mathrm{~nm}$
AP EAPCET-12.07.2022
ATOMS
145521
Let $\lambda_{L}$ and $\lambda_{s}$ be longest wavelength photon and shortest wavelength photon respectively in the Balmer series. The value of $\frac{\lambda_{L}}{\lambda_{s}}$ is
1 $\frac{9}{5}$
2 2
3 $\frac{5}{2}$
4 $\frac{8}{3}$
Explanation:
A We know that, For Balmer series $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ For shortest wavelength $\mathrm{n}=2 \rightarrow \mathrm{n}=\infty$ For longest wavelength $\mathrm{n}=2 \rightarrow \mathrm{n}=3$ $\frac{1}{\lambda_{\mathrm{s}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)=\frac{\mathrm{R}}{4}$ $\lambda_{\mathrm{s}}=\frac{4}{\mathrm{R}}$ And, $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \Rightarrow \mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36} \mathrm{R}$ $\lambda_{\mathrm{L}}=\frac{36}{5 \mathrm{R}}$ Then ratio, $\frac{\lambda_{\mathrm{L}}}{\lambda_{\mathrm{S}}}=\frac{\frac{36}{5 \mathrm{R}}}{\frac{4}{\mathrm{R}}}=\frac{9}{5}$ $\frac{\lambda_{\mathrm{L}}}{\lambda_{\mathrm{S}}}=\frac{9}{5}$
AP EAMCET-11.07.2022
ATOMS
145522
Energy of a stationary electron in the hydrogen atom is $E=-\frac{13.6}{n^{2}} \mathrm{eV}$ then the energies required to excite the electron in hydrogen atom to (a) its second excited state and (b) ionized state respectively
1 $\sim 10 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
2 $\sim 12 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
3 $\sim 12 \mathrm{eV}$, (b) $10.6 \mathrm{eV}$
4 $\sim 8 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
Explanation:
B Energy of stationary, $E=\frac{-13.6}{n^{2}} \mathrm{eV}$ Then energy of $\mathrm{e}^{-}$in second excited state, $\mathrm{n}=3$ $\mathrm{E}=\frac{-13.6}{(3)^{2}}=\frac{-13.6}{9}=-1.51 \mathrm{eV}$ Then required energy $=\mathrm{E}_{\text {excited }}-\mathrm{E}_{\text {ground }}$ $=(-1.51)-(-13.6)$ $=13.6-1.51$ $\mathrm{E}=12.09 \mathrm{eV}$ Then, $\quad \mathrm{E} \square 12 \mathrm{eV}$ And ionization state energy will be $13.6 \mathrm{eV}$.
AP EAMCET-04.07.2022
ATOMS
145524
Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
1 $3: 4$
2 $4: 3$
3 $1: 4$
4 $4: 1$
Explanation:
A Energy of photon produced as the result of electron transition $\mathrm{E}_{\mathrm{n}_{1} \rightarrow \mathrm{n}_{2}}=\operatorname{Rhc}\left(\frac{1}{\mathrm{n}_{\mathrm{f}}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ Then energy of photon when electron transition $\mathrm{n}_{\mathrm{i}}=2 \text { to } \mathrm{n}_{\mathrm{f}}=1$ $\mathrm{E}_{2 \rightarrow 1}=\operatorname{Rhc}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=\frac{3}{4} \mathrm{Rhc}$ But when electron transition form highest energy level to first energy level then, $\mathrm{n}_{\mathrm{i}}=\infty \text { to } \mathrm{n}_{\mathrm{f}}=1$ $\mathrm{E}_{\infty \rightarrow 1}=\operatorname{Rhc}\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right)=\text { Rhc }$ Then ratio, $\frac{E_{2 \rightarrow 1}}{E_{\infty \rightarrow 1}}=\frac{\frac{3}{4} R h c}{\text { Rhc }}=\frac{3}{4}$ $\frac{E_{2 \rightarrow 1}}{E_{\infty \rightarrow 1}}=\frac{3}{4}$ $E_{2}: E_{\infty}=3: 4$
NEET Test Series from KOTA - 10 Papers In MS WORD
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ATOMS
145520
The shortest wavelength in the Balmer series of hydrogen atom spectrum is [For hydrogen, Rydberg constant = $1.097 \times 10^{7} \mathrm{~m}^{-1}$ ]
1 $91.2 \mathrm{~nm}$
2 $364.6 \mathrm{~nm}$
3 $820.4 \mathrm{~nm}$
4 $2278.9 \mathrm{~nm}$
Explanation:
B We know that, $\mathrm{R}=1.097 \times 10^{7} \mathrm{~m}^{-1}$ The shortest wavelength of Balmer series $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda}=\frac{1.097 \times 10^{7}}{4}$ $\lambda=\frac{4}{1.097} \times 10^{-7} \mathrm{~m}$ $\lambda=3.6463 \times 10^{-7}=364.6 \times 10^{-9} \mathrm{~m}$ $\lambda=364.6 \mathrm{~nm}$
AP EAPCET-12.07.2022
ATOMS
145521
Let $\lambda_{L}$ and $\lambda_{s}$ be longest wavelength photon and shortest wavelength photon respectively in the Balmer series. The value of $\frac{\lambda_{L}}{\lambda_{s}}$ is
1 $\frac{9}{5}$
2 2
3 $\frac{5}{2}$
4 $\frac{8}{3}$
Explanation:
A We know that, For Balmer series $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ For shortest wavelength $\mathrm{n}=2 \rightarrow \mathrm{n}=\infty$ For longest wavelength $\mathrm{n}=2 \rightarrow \mathrm{n}=3$ $\frac{1}{\lambda_{\mathrm{s}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)=\frac{\mathrm{R}}{4}$ $\lambda_{\mathrm{s}}=\frac{4}{\mathrm{R}}$ And, $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \Rightarrow \mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36} \mathrm{R}$ $\lambda_{\mathrm{L}}=\frac{36}{5 \mathrm{R}}$ Then ratio, $\frac{\lambda_{\mathrm{L}}}{\lambda_{\mathrm{S}}}=\frac{\frac{36}{5 \mathrm{R}}}{\frac{4}{\mathrm{R}}}=\frac{9}{5}$ $\frac{\lambda_{\mathrm{L}}}{\lambda_{\mathrm{S}}}=\frac{9}{5}$
AP EAMCET-11.07.2022
ATOMS
145522
Energy of a stationary electron in the hydrogen atom is $E=-\frac{13.6}{n^{2}} \mathrm{eV}$ then the energies required to excite the electron in hydrogen atom to (a) its second excited state and (b) ionized state respectively
1 $\sim 10 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
2 $\sim 12 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
3 $\sim 12 \mathrm{eV}$, (b) $10.6 \mathrm{eV}$
4 $\sim 8 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
Explanation:
B Energy of stationary, $E=\frac{-13.6}{n^{2}} \mathrm{eV}$ Then energy of $\mathrm{e}^{-}$in second excited state, $\mathrm{n}=3$ $\mathrm{E}=\frac{-13.6}{(3)^{2}}=\frac{-13.6}{9}=-1.51 \mathrm{eV}$ Then required energy $=\mathrm{E}_{\text {excited }}-\mathrm{E}_{\text {ground }}$ $=(-1.51)-(-13.6)$ $=13.6-1.51$ $\mathrm{E}=12.09 \mathrm{eV}$ Then, $\quad \mathrm{E} \square 12 \mathrm{eV}$ And ionization state energy will be $13.6 \mathrm{eV}$.
AP EAMCET-04.07.2022
ATOMS
145524
Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
1 $3: 4$
2 $4: 3$
3 $1: 4$
4 $4: 1$
Explanation:
A Energy of photon produced as the result of electron transition $\mathrm{E}_{\mathrm{n}_{1} \rightarrow \mathrm{n}_{2}}=\operatorname{Rhc}\left(\frac{1}{\mathrm{n}_{\mathrm{f}}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ Then energy of photon when electron transition $\mathrm{n}_{\mathrm{i}}=2 \text { to } \mathrm{n}_{\mathrm{f}}=1$ $\mathrm{E}_{2 \rightarrow 1}=\operatorname{Rhc}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=\frac{3}{4} \mathrm{Rhc}$ But when electron transition form highest energy level to first energy level then, $\mathrm{n}_{\mathrm{i}}=\infty \text { to } \mathrm{n}_{\mathrm{f}}=1$ $\mathrm{E}_{\infty \rightarrow 1}=\operatorname{Rhc}\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right)=\text { Rhc }$ Then ratio, $\frac{E_{2 \rightarrow 1}}{E_{\infty \rightarrow 1}}=\frac{\frac{3}{4} R h c}{\text { Rhc }}=\frac{3}{4}$ $\frac{E_{2 \rightarrow 1}}{E_{\infty \rightarrow 1}}=\frac{3}{4}$ $E_{2}: E_{\infty}=3: 4$
145520
The shortest wavelength in the Balmer series of hydrogen atom spectrum is [For hydrogen, Rydberg constant = $1.097 \times 10^{7} \mathrm{~m}^{-1}$ ]
1 $91.2 \mathrm{~nm}$
2 $364.6 \mathrm{~nm}$
3 $820.4 \mathrm{~nm}$
4 $2278.9 \mathrm{~nm}$
Explanation:
B We know that, $\mathrm{R}=1.097 \times 10^{7} \mathrm{~m}^{-1}$ The shortest wavelength of Balmer series $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda}=\frac{1.097 \times 10^{7}}{4}$ $\lambda=\frac{4}{1.097} \times 10^{-7} \mathrm{~m}$ $\lambda=3.6463 \times 10^{-7}=364.6 \times 10^{-9} \mathrm{~m}$ $\lambda=364.6 \mathrm{~nm}$
AP EAPCET-12.07.2022
ATOMS
145521
Let $\lambda_{L}$ and $\lambda_{s}$ be longest wavelength photon and shortest wavelength photon respectively in the Balmer series. The value of $\frac{\lambda_{L}}{\lambda_{s}}$ is
1 $\frac{9}{5}$
2 2
3 $\frac{5}{2}$
4 $\frac{8}{3}$
Explanation:
A We know that, For Balmer series $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ For shortest wavelength $\mathrm{n}=2 \rightarrow \mathrm{n}=\infty$ For longest wavelength $\mathrm{n}=2 \rightarrow \mathrm{n}=3$ $\frac{1}{\lambda_{\mathrm{s}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)=\frac{\mathrm{R}}{4}$ $\lambda_{\mathrm{s}}=\frac{4}{\mathrm{R}}$ And, $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \Rightarrow \mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36} \mathrm{R}$ $\lambda_{\mathrm{L}}=\frac{36}{5 \mathrm{R}}$ Then ratio, $\frac{\lambda_{\mathrm{L}}}{\lambda_{\mathrm{S}}}=\frac{\frac{36}{5 \mathrm{R}}}{\frac{4}{\mathrm{R}}}=\frac{9}{5}$ $\frac{\lambda_{\mathrm{L}}}{\lambda_{\mathrm{S}}}=\frac{9}{5}$
AP EAMCET-11.07.2022
ATOMS
145522
Energy of a stationary electron in the hydrogen atom is $E=-\frac{13.6}{n^{2}} \mathrm{eV}$ then the energies required to excite the electron in hydrogen atom to (a) its second excited state and (b) ionized state respectively
1 $\sim 10 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
2 $\sim 12 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
3 $\sim 12 \mathrm{eV}$, (b) $10.6 \mathrm{eV}$
4 $\sim 8 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
Explanation:
B Energy of stationary, $E=\frac{-13.6}{n^{2}} \mathrm{eV}$ Then energy of $\mathrm{e}^{-}$in second excited state, $\mathrm{n}=3$ $\mathrm{E}=\frac{-13.6}{(3)^{2}}=\frac{-13.6}{9}=-1.51 \mathrm{eV}$ Then required energy $=\mathrm{E}_{\text {excited }}-\mathrm{E}_{\text {ground }}$ $=(-1.51)-(-13.6)$ $=13.6-1.51$ $\mathrm{E}=12.09 \mathrm{eV}$ Then, $\quad \mathrm{E} \square 12 \mathrm{eV}$ And ionization state energy will be $13.6 \mathrm{eV}$.
AP EAMCET-04.07.2022
ATOMS
145524
Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
1 $3: 4$
2 $4: 3$
3 $1: 4$
4 $4: 1$
Explanation:
A Energy of photon produced as the result of electron transition $\mathrm{E}_{\mathrm{n}_{1} \rightarrow \mathrm{n}_{2}}=\operatorname{Rhc}\left(\frac{1}{\mathrm{n}_{\mathrm{f}}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ Then energy of photon when electron transition $\mathrm{n}_{\mathrm{i}}=2 \text { to } \mathrm{n}_{\mathrm{f}}=1$ $\mathrm{E}_{2 \rightarrow 1}=\operatorname{Rhc}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=\frac{3}{4} \mathrm{Rhc}$ But when electron transition form highest energy level to first energy level then, $\mathrm{n}_{\mathrm{i}}=\infty \text { to } \mathrm{n}_{\mathrm{f}}=1$ $\mathrm{E}_{\infty \rightarrow 1}=\operatorname{Rhc}\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right)=\text { Rhc }$ Then ratio, $\frac{E_{2 \rightarrow 1}}{E_{\infty \rightarrow 1}}=\frac{\frac{3}{4} R h c}{\text { Rhc }}=\frac{3}{4}$ $\frac{E_{2 \rightarrow 1}}{E_{\infty \rightarrow 1}}=\frac{3}{4}$ $E_{2}: E_{\infty}=3: 4$
145520
The shortest wavelength in the Balmer series of hydrogen atom spectrum is [For hydrogen, Rydberg constant = $1.097 \times 10^{7} \mathrm{~m}^{-1}$ ]
1 $91.2 \mathrm{~nm}$
2 $364.6 \mathrm{~nm}$
3 $820.4 \mathrm{~nm}$
4 $2278.9 \mathrm{~nm}$
Explanation:
B We know that, $\mathrm{R}=1.097 \times 10^{7} \mathrm{~m}^{-1}$ The shortest wavelength of Balmer series $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda}=\frac{1.097 \times 10^{7}}{4}$ $\lambda=\frac{4}{1.097} \times 10^{-7} \mathrm{~m}$ $\lambda=3.6463 \times 10^{-7}=364.6 \times 10^{-9} \mathrm{~m}$ $\lambda=364.6 \mathrm{~nm}$
AP EAPCET-12.07.2022
ATOMS
145521
Let $\lambda_{L}$ and $\lambda_{s}$ be longest wavelength photon and shortest wavelength photon respectively in the Balmer series. The value of $\frac{\lambda_{L}}{\lambda_{s}}$ is
1 $\frac{9}{5}$
2 2
3 $\frac{5}{2}$
4 $\frac{8}{3}$
Explanation:
A We know that, For Balmer series $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ For shortest wavelength $\mathrm{n}=2 \rightarrow \mathrm{n}=\infty$ For longest wavelength $\mathrm{n}=2 \rightarrow \mathrm{n}=3$ $\frac{1}{\lambda_{\mathrm{s}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)=\frac{\mathrm{R}}{4}$ $\lambda_{\mathrm{s}}=\frac{4}{\mathrm{R}}$ And, $\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \Rightarrow \mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36} \mathrm{R}$ $\lambda_{\mathrm{L}}=\frac{36}{5 \mathrm{R}}$ Then ratio, $\frac{\lambda_{\mathrm{L}}}{\lambda_{\mathrm{S}}}=\frac{\frac{36}{5 \mathrm{R}}}{\frac{4}{\mathrm{R}}}=\frac{9}{5}$ $\frac{\lambda_{\mathrm{L}}}{\lambda_{\mathrm{S}}}=\frac{9}{5}$
AP EAMCET-11.07.2022
ATOMS
145522
Energy of a stationary electron in the hydrogen atom is $E=-\frac{13.6}{n^{2}} \mathrm{eV}$ then the energies required to excite the electron in hydrogen atom to (a) its second excited state and (b) ionized state respectively
1 $\sim 10 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
2 $\sim 12 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
3 $\sim 12 \mathrm{eV}$, (b) $10.6 \mathrm{eV}$
4 $\sim 8 \mathrm{eV}$, (b) $13.6 \mathrm{eV}$
Explanation:
B Energy of stationary, $E=\frac{-13.6}{n^{2}} \mathrm{eV}$ Then energy of $\mathrm{e}^{-}$in second excited state, $\mathrm{n}=3$ $\mathrm{E}=\frac{-13.6}{(3)^{2}}=\frac{-13.6}{9}=-1.51 \mathrm{eV}$ Then required energy $=\mathrm{E}_{\text {excited }}-\mathrm{E}_{\text {ground }}$ $=(-1.51)-(-13.6)$ $=13.6-1.51$ $\mathrm{E}=12.09 \mathrm{eV}$ Then, $\quad \mathrm{E} \square 12 \mathrm{eV}$ And ionization state energy will be $13.6 \mathrm{eV}$.
AP EAMCET-04.07.2022
ATOMS
145524
Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level.
1 $3: 4$
2 $4: 3$
3 $1: 4$
4 $4: 1$
Explanation:
A Energy of photon produced as the result of electron transition $\mathrm{E}_{\mathrm{n}_{1} \rightarrow \mathrm{n}_{2}}=\operatorname{Rhc}\left(\frac{1}{\mathrm{n}_{\mathrm{f}}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ Then energy of photon when electron transition $\mathrm{n}_{\mathrm{i}}=2 \text { to } \mathrm{n}_{\mathrm{f}}=1$ $\mathrm{E}_{2 \rightarrow 1}=\operatorname{Rhc}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)=\frac{3}{4} \mathrm{Rhc}$ But when electron transition form highest energy level to first energy level then, $\mathrm{n}_{\mathrm{i}}=\infty \text { to } \mathrm{n}_{\mathrm{f}}=1$ $\mathrm{E}_{\infty \rightarrow 1}=\operatorname{Rhc}\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right)=\text { Rhc }$ Then ratio, $\frac{E_{2 \rightarrow 1}}{E_{\infty \rightarrow 1}}=\frac{\frac{3}{4} R h c}{\text { Rhc }}=\frac{3}{4}$ $\frac{E_{2 \rightarrow 1}}{E_{\infty \rightarrow 1}}=\frac{3}{4}$ $E_{2}: E_{\infty}=3: 4$