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For balmer series in the spectrum of a...

For balmer series in the spectrum of atomic hydrogen the wave number of each line is given by
`bar(V)=R_(H)(1)/(n_(1)^(2))-(1)/(n_(2)^(2))` where `R_(H)` is a constant and `n_(1)` and `n_(2)` are integers which of the following statement (S) is / are correct
1 as wavelength decreases the lines in the series converge
2 The integer `n_(1)` is equal to 2
3 The ionization energy of hydrogen can be calculated from the wave number of these lines
4 The line of longest wavelength corresponds to `n_(2)=3`

A

1,2 and 3

B

2,3 and 4

C

1,2 and 4

D

2 and 4 only

Text Solution

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The correct Answer is:
To solve the question regarding the Balmer series in the spectrum of atomic hydrogen, we will analyze each statement provided in the question step by step. ### Given: The wave number of each line in the Balmer series is given by: \[ \bar{V} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R_H \) is a constant, and \( n_1 \) and \( n_2 \) are integers. ### Analyzing Each Statement: 1. **Statement 1**: As wavelength decreases, the lines in the series converge. - **Explanation**: In the context of wave numbers, as the wavelength decreases, the wave number \( \bar{V} \) increases. For the Balmer series, \( n_1 \) is fixed at 2, while \( n_2 \) can take values 3, 4, 5, etc. As \( n_2 \) increases, the difference \( \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) becomes smaller, leading to convergence of the lines. Thus, this statement is **True**. 2. **Statement 2**: The integer \( n_1 \) is equal to 2. - **Explanation**: In the Balmer series, transitions occur from higher energy levels \( n_2 \) (3, 4, 5, ...) to \( n_1 = 2 \). Therefore, this statement is **True**. 3. **Statement 3**: The ionization energy of hydrogen can be calculated from the wave number of these lines. - **Explanation**: The ionization energy refers to the energy required to remove an electron completely from the atom. The Balmer series only involves transitions where \( n_1 = 2 \) and \( n_2 \) is greater than 2. To calculate ionization energy, we would need to consider transitions from \( n = 1 \) to \( n = \infty \). Therefore, this statement is **False**. 4. **Statement 4**: The line of longest wavelength corresponds to \( n_2 = 3 \). - **Explanation**: The longest wavelength corresponds to the smallest energy transition, which occurs when \( n_2 \) is at its minimum value. For the Balmer series, the transition from \( n_2 = 3 \) to \( n_1 = 2 \) gives the longest wavelength. Therefore, this statement is **True**. ### Conclusion: Based on the analysis: - Statement 1: True - Statement 2: True - Statement 3: False - Statement 4: True Thus, the correct statements are 1, 2, and 4. ### Final Answer: The correct option is **1, 2, and 4**. ---
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For balmer series in the spectrum of atomic hydrogen the wave number of each line is given by bar(V) = R[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))] where R_(H) is a constant and n_(1) and n_(2) are integers. Which of the following statements (s), is (are correct) 1. As wave length decreases the lines in the series converge 2. The integer n_(1) is equal to 2. 3. The ionisation energy of hydrogen can be calculated from the wave numbers of three lines. 4. The line of shortest wavelength corresponds to = 3.

In hydrogen spectrum wave number of different lines is given by 1/lambda=R_(H)[1/n_(i)^(2)-1/n_(f)^(2)] where R_(H)=1.090678xx10^(7)m^(-1) The wavelength of first line of Lyman series would be

For the Paschen series thr values of n_(1) and n_(2) in the expression Delta E = R_(H)c [(1)/(n_(1)^(2))-(1)/(n_(2)^(2))] are

the only electron in the hydrogen atom resides under ordinary conditions in the first orbit. When energy is supplied the electron moves to higher energy orbit depending on the amount of energy absorbed. It emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second. Similarly, Paschen, Breakett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively (as shown in figure). Maximum number of different lines produced when electron jump from nth level to ground level is equal to (n(n-1))/(2). For example in the case of n=4, number of lines produced is 6.(4to3,4to2,4to1,3to2,3to1,2to1). When an electron returns from n_(2) "to"n_(1) state, the number of different lines in the spectrum will be equal to ((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2) If the electron comes back from energy level having energy E_(2) to energy level having energy E_(1), then the difference may be expressed in terms of energy of photon as: E_(2)-E_(1)=DeltaE,lambda=(hc)/(DeltaE),DeltaE=hv(v-"frequency") Since h and c are constants DeltaE corresponds to definite energy: thus each transition from one energy level to another will produce a light of definite wavelength. This is actually observed as a line in the spectrum of hydrogen atom. Wave number of line is given by the formula: barv=RZ^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) where R is a Rydberg constant (R=1.1xx10^(7)m^(-1)). (i) First line of a series : It is called line of longest wavelength of line of smallest energy'. (ii) Series limit or last line of a series : It is the line of shortest wavelength or line of highest energy. Last line of breakett series for H-atom has wavelength lambda_(1)"Å" and 2nd line of Lyman series has wavelength lambda_(2)"Å" then:

the only electron in the hydrogen atom resides under ordinary conditions in the first orbit. When energy is supplied the electron moves to higher energy orbit depending on the amount of energy absorbed. It emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second. Similarly, Paschen, Breakett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively (as shown in figure). Maximum number of different lines produced when electron jump from nth level to ground level is equal to (n(n-1))/(2). For example in the case of n=4, number of lines produced is 6.(4to3,4to2,4to1,3to2,3to1,2to1). When an electron returns from n_(2) "to"n_(1) state, the number of different lines in the spectrum will be equal to ((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2) If the electron comes back from energy level having energy E_(2) to energy level having energy E_(1), then the difference may be expressed in terms of energy of photon as: E_(2)-E_(1)=DeltaE,lambda=(hc)/(DeltaE),DeltaE=hv(v-"frequency") Since h and c are constants DeltaE corresponds to definite energy: thus each transition from one energy level to another will produce a light of definite wavelength. This is actually observed as a line in the spectrum of hydrogen atom. Wave number of line is given by the formula: barv=RZ^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) where R is a Rydberg constant (R=1.1xx10^(7)m^(-1)). (i) First line of a series : It is called line of longest wavelength of line of smallest energy'. (ii) Series limit or last line of a series : It is the line of shortest wavelength or line of highest energy. wave number of the first line of Paschen series in Be^(3+) ion is :

the only electron in the hydrogen atom resides under ordinary conditions in the first orbit. When energy is supplied the electron moves to higher energy orbit depending on the amount of energy absorbed. It emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second. Similarly, Paschen, Breakett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively (as shown in figure). Maximum number of different lines produced when electron jump from nth level to ground level is equal to (n(n-1))/(2). For example in the case of n=4, number of lines produced is 6.(4to3,4to2,4to1,3to2,3to1,2to1). When an electron returns from n_(2) "to"n_(1) state, the number of different lines in the spectrum will be equal to ((n_(2)-n_(1))(n_(2)-n_(1)+1))/(2) If the electron comes back from energy level having energy E_(2) to energy level having energy E_(1), then the difference may be expressed in terms of energy of photon as: E_(2)-E_(1)=DeltaE,lambda=(hc)/(DeltaE),DeltaE=hv(v-"frequency") Since h and c are constants DeltaE corresponds to definite energy: thus each transition from one energy level to another will produce a light of definite wavelength. This is actually observed as a line in the spectrum of hydrogen atom. Wave number of line is given by the formula: barv=RZ^(2)((1)/(n_(1)^(2))-(1)/(n_(2)^(2))) where R is a Rydberg constant (R=1.1xx10^(7)m^(-1)). (i) First line of a series : It is called line of longest wavelength of line of smallest energy'. (ii) Series limit or last line of a series : It is the line of shortest wavelength or line of highest energy. Consider the following statements (i) Spectral lines of He^(+) ion belonging to Balmer series are not in visible range. (ii) In the Balmer series of H-atom maximum lines are in ultraviolet ragion. (iii) 2nd line of Lyman series of He^(+) ion has energy 48.35 eV The above statement (i), (ii), (iii) respectively are (T=True, F=False):

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