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Last line of brackett series for H atom ...

Last line of brackett series for H atom has `lambda_(1) A wavelength 2^(nd)` line of lyman series has wavelength `lambda_(2) A` then

A

`(128)/(lambda_(1))=(6)/(lambda_(2))`

B

`(16)/(lambda_(1))=(9)/(8lambda_(2))`

C

`(4)/(lambda_(1))=(1)/(lambda_(2))`

D

`(128)/(lambda_(1))=(8)/(lambda_(2))`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the relationship between the wavelengths of the last line of the Brackett series and the second line of the Lyman series for the hydrogen atom. ### Step-by-Step Solution: 1. **Identify the Series and Transitions**: - The last line of the Brackett series corresponds to a transition from \( n_2 \) to \( n_1 \) where \( n_1 = 4 \) and \( n_2 \) approaches infinity (\( n_2 \to \infty \)). - The second line of the Lyman series corresponds to a transition from \( n_2 = 3 \) to \( n_1 = 1 \). 2. **Use the Rydberg Formula**: The Rydberg formula for the wavelength of emitted light during an electron transition is given by: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R_H \) is the Rydberg constant. 3. **Calculate Wavelength for Brackett Series**: - For the last line of the Brackett series: \[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{16} - 0 \right) = \frac{R_H}{16} \] - Therefore, we can express \( \lambda_1 \): \[ \lambda_1 = \frac{16}{R_H} \] 4. **Calculate Wavelength for Lyman Series**: - For the second line of the Lyman series: \[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R_H \left( 1 - \frac{1}{9} \right) = R_H \left( \frac{8}{9} \right) \] - Therefore, we can express \( \lambda_2 \): \[ \lambda_2 = \frac{9}{8R_H} \] 5. **Find the Ratio of Wavelengths**: - Now, we can find the ratio \( \frac{\lambda_1}{\lambda_2} \): \[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{16}{R_H}}{\frac{9}{8R_H}} = \frac{16 \cdot 8}{9} = \frac{128}{9} \] 6. **Conclusion**: - The relationship between the wavelengths is: \[ \frac{\lambda_1}{\lambda_2} = \frac{128}{9} \] ### Final Answer: The correct relationship is \( \lambda_1 : \lambda_2 = 128 : 9 \).
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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:

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