145513
In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Bracket series is:
1 $16 \lambda$
2 $2 \lambda$
3 $4 \lambda$
4 $9 \lambda$
Explanation:
C Formula for the shortest wavelength line that of Balmer series is given as $\mathrm{n}_{1}=2, \mathrm{n}_{2}=\infty$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $=\mathrm{R}\left[\frac{1}{2^{2}}-\frac{1}{\infty}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{4}\right]$ $\lambda=\frac{4}{\mathrm{R}}$ For Bracket series - $\mathrm{n}_{1}=4 \text { and } \mathrm{n}_{2}=\infty$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda_{2}}=\frac{\mathrm{R}}{16}$ $\frac{1}{\lambda_{2}}=\frac{4}{\lambda .16}$ $\lambda_{2}=4 \lambda$
NEET (UG) 07.05.2023
ATOMS
145516
The energy levels of an atom is shown in figure. Which one of these transitions will result in the emission of a photon of wavelength $124.1 \mathrm{~nm}$ ? Given $\left(h=6.62 \times 10^{-34} \mathrm{Js}\right)$
1 D
2 $\mathrm{B}$
3 $\mathrm{A}$
4 $\mathrm{C}$
Explanation:
A Given that, $\lambda=124.1 \mathrm{~nm}=124.1 \times 10^{-9}$ $\mathrm{h}=6.62 \times 10^{-34} \mathrm{Js}$ We know that, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{124.1 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{eV}$ $\mathrm{E}=\frac{12.41 \times 10^{2}}{124.1} \mathrm{eV}$ $\mathrm{E}=10 \mathrm{eV}$ Hence, transition D will result in emission of photon.
JEE Main-25.01.2023
ATOMS
145517
A photon is emitted in transition from $n=4$ to $\mathrm{n}=1$ level in hydrogen atom. The corresponding wavelength for this transition is (given, $\mathrm{h}=4 \times 10^{-15} \mathrm{eVs}$ ):
145513
In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Bracket series is:
1 $16 \lambda$
2 $2 \lambda$
3 $4 \lambda$
4 $9 \lambda$
Explanation:
C Formula for the shortest wavelength line that of Balmer series is given as $\mathrm{n}_{1}=2, \mathrm{n}_{2}=\infty$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $=\mathrm{R}\left[\frac{1}{2^{2}}-\frac{1}{\infty}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{4}\right]$ $\lambda=\frac{4}{\mathrm{R}}$ For Bracket series - $\mathrm{n}_{1}=4 \text { and } \mathrm{n}_{2}=\infty$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda_{2}}=\frac{\mathrm{R}}{16}$ $\frac{1}{\lambda_{2}}=\frac{4}{\lambda .16}$ $\lambda_{2}=4 \lambda$
NEET (UG) 07.05.2023
ATOMS
145516
The energy levels of an atom is shown in figure. Which one of these transitions will result in the emission of a photon of wavelength $124.1 \mathrm{~nm}$ ? Given $\left(h=6.62 \times 10^{-34} \mathrm{Js}\right)$
1 D
2 $\mathrm{B}$
3 $\mathrm{A}$
4 $\mathrm{C}$
Explanation:
A Given that, $\lambda=124.1 \mathrm{~nm}=124.1 \times 10^{-9}$ $\mathrm{h}=6.62 \times 10^{-34} \mathrm{Js}$ We know that, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{124.1 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{eV}$ $\mathrm{E}=\frac{12.41 \times 10^{2}}{124.1} \mathrm{eV}$ $\mathrm{E}=10 \mathrm{eV}$ Hence, transition D will result in emission of photon.
JEE Main-25.01.2023
ATOMS
145517
A photon is emitted in transition from $n=4$ to $\mathrm{n}=1$ level in hydrogen atom. The corresponding wavelength for this transition is (given, $\mathrm{h}=4 \times 10^{-15} \mathrm{eVs}$ ):
145513
In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Bracket series is:
1 $16 \lambda$
2 $2 \lambda$
3 $4 \lambda$
4 $9 \lambda$
Explanation:
C Formula for the shortest wavelength line that of Balmer series is given as $\mathrm{n}_{1}=2, \mathrm{n}_{2}=\infty$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $=\mathrm{R}\left[\frac{1}{2^{2}}-\frac{1}{\infty}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{4}\right]$ $\lambda=\frac{4}{\mathrm{R}}$ For Bracket series - $\mathrm{n}_{1}=4 \text { and } \mathrm{n}_{2}=\infty$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda_{2}}=\frac{\mathrm{R}}{16}$ $\frac{1}{\lambda_{2}}=\frac{4}{\lambda .16}$ $\lambda_{2}=4 \lambda$
NEET (UG) 07.05.2023
ATOMS
145516
The energy levels of an atom is shown in figure. Which one of these transitions will result in the emission of a photon of wavelength $124.1 \mathrm{~nm}$ ? Given $\left(h=6.62 \times 10^{-34} \mathrm{Js}\right)$
1 D
2 $\mathrm{B}$
3 $\mathrm{A}$
4 $\mathrm{C}$
Explanation:
A Given that, $\lambda=124.1 \mathrm{~nm}=124.1 \times 10^{-9}$ $\mathrm{h}=6.62 \times 10^{-34} \mathrm{Js}$ We know that, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{124.1 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{eV}$ $\mathrm{E}=\frac{12.41 \times 10^{2}}{124.1} \mathrm{eV}$ $\mathrm{E}=10 \mathrm{eV}$ Hence, transition D will result in emission of photon.
JEE Main-25.01.2023
ATOMS
145517
A photon is emitted in transition from $n=4$ to $\mathrm{n}=1$ level in hydrogen atom. The corresponding wavelength for this transition is (given, $\mathrm{h}=4 \times 10^{-15} \mathrm{eVs}$ ):
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
ATOMS
145513
In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Bracket series is:
1 $16 \lambda$
2 $2 \lambda$
3 $4 \lambda$
4 $9 \lambda$
Explanation:
C Formula for the shortest wavelength line that of Balmer series is given as $\mathrm{n}_{1}=2, \mathrm{n}_{2}=\infty$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $=\mathrm{R}\left[\frac{1}{2^{2}}-\frac{1}{\infty}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{4}\right]$ $\lambda=\frac{4}{\mathrm{R}}$ For Bracket series - $\mathrm{n}_{1}=4 \text { and } \mathrm{n}_{2}=\infty$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda_{2}}=\frac{\mathrm{R}}{16}$ $\frac{1}{\lambda_{2}}=\frac{4}{\lambda .16}$ $\lambda_{2}=4 \lambda$
NEET (UG) 07.05.2023
ATOMS
145516
The energy levels of an atom is shown in figure. Which one of these transitions will result in the emission of a photon of wavelength $124.1 \mathrm{~nm}$ ? Given $\left(h=6.62 \times 10^{-34} \mathrm{Js}\right)$
1 D
2 $\mathrm{B}$
3 $\mathrm{A}$
4 $\mathrm{C}$
Explanation:
A Given that, $\lambda=124.1 \mathrm{~nm}=124.1 \times 10^{-9}$ $\mathrm{h}=6.62 \times 10^{-34} \mathrm{Js}$ We know that, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{124.1 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{eV}$ $\mathrm{E}=\frac{12.41 \times 10^{2}}{124.1} \mathrm{eV}$ $\mathrm{E}=10 \mathrm{eV}$ Hence, transition D will result in emission of photon.
JEE Main-25.01.2023
ATOMS
145517
A photon is emitted in transition from $n=4$ to $\mathrm{n}=1$ level in hydrogen atom. The corresponding wavelength for this transition is (given, $\mathrm{h}=4 \times 10^{-15} \mathrm{eVs}$ ):
145513
In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Bracket series is:
1 $16 \lambda$
2 $2 \lambda$
3 $4 \lambda$
4 $9 \lambda$
Explanation:
C Formula for the shortest wavelength line that of Balmer series is given as $\mathrm{n}_{1}=2, \mathrm{n}_{2}=\infty$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $=\mathrm{R}\left[\frac{1}{2^{2}}-\frac{1}{\infty}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{4}\right]$ $\lambda=\frac{4}{\mathrm{R}}$ For Bracket series - $\mathrm{n}_{1}=4 \text { and } \mathrm{n}_{2}=\infty$ $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda_{2}}=\frac{\mathrm{R}}{16}$ $\frac{1}{\lambda_{2}}=\frac{4}{\lambda .16}$ $\lambda_{2}=4 \lambda$
NEET (UG) 07.05.2023
ATOMS
145516
The energy levels of an atom is shown in figure. Which one of these transitions will result in the emission of a photon of wavelength $124.1 \mathrm{~nm}$ ? Given $\left(h=6.62 \times 10^{-34} \mathrm{Js}\right)$
1 D
2 $\mathrm{B}$
3 $\mathrm{A}$
4 $\mathrm{C}$
Explanation:
A Given that, $\lambda=124.1 \mathrm{~nm}=124.1 \times 10^{-9}$ $\mathrm{h}=6.62 \times 10^{-34} \mathrm{Js}$ We know that, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}=\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{124.1 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{eV}$ $\mathrm{E}=\frac{12.41 \times 10^{2}}{124.1} \mathrm{eV}$ $\mathrm{E}=10 \mathrm{eV}$ Hence, transition D will result in emission of photon.
JEE Main-25.01.2023
ATOMS
145517
A photon is emitted in transition from $n=4$ to $\mathrm{n}=1$ level in hydrogen atom. The corresponding wavelength for this transition is (given, $\mathrm{h}=4 \times 10^{-15} \mathrm{eVs}$ ):
