Home
Class 12
CHEMISTRY
The energy of n^(th) orbit is given by ...

The energy of `n^(th)` orbit is given by
`E_(n) = ( -Rhc)/(n^(2))`
When electron jumpsfrom one orbit to another orbit then wavelength associated with the radiation is given by
`(1)/(lambda) = RZ^(2)((1)/(n_(1)^(2)) - (1)/ (n_(2)^(2)))`
The series that belongs to visible region is

A

Lyman Series

B

Balmer Series

C

Pfund Series

D

Humphrey Series

Text Solution

AI Generated Solution

The correct Answer is:
To determine which series belongs to the visible region, we can follow these steps: ### Step 1: Understand the Energy Equation The energy of the nth orbit is given by: \[ E_n = \frac{-Rhc}{n^2} \] where \( R \) is the Rydberg constant, \( h \) is Planck's constant, and \( c \) is the speed of light. ### Step 2: Wavelength Calculation When an electron jumps from one orbit to another, the wavelength associated with the radiation emitted or absorbed is given by: \[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( Z \) is the atomic number, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level. ### Step 3: Identify the Visible Region The visible region of the electromagnetic spectrum ranges from approximately 380 nm to 700 nm. ### Step 4: Identify the Series - **Lyman series**: Transitions to \( n=1 \) (UV region) - **Balmer series**: Transitions to \( n=2 \) (Visible region) - **Paschen series**: Transitions to \( n=3 \) (IR region) - **Humphreys series**: Transitions to \( n=4 \) and higher (far IR region) ### Step 5: Determine the Balmer Series The Balmer series corresponds to transitions where \( n_1 = 2 \) and \( n_2 \) can be 3, 4, 5, etc. ### Step 6: Calculate Wavelengths for Balmer Series 1. For \( n_2 = 3 \) and \( n_1 = 2 \): \[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] Calculate \( \lambda \) to find the wavelength. 2. For \( n_2 = 4 \) and \( n_1 = 2 \): \[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \] Again, calculate \( \lambda \). ### Step 7: Compare Wavelengths After calculating the wavelengths for the transitions in the Balmer series, check if they fall within the visible range (380 nm to 700 nm). ### Conclusion The series that belongs to the visible region is the **Balmer series**. ---
Promotional Banner

Topper's Solved these Questions

  • STRUCTURE OF ATOM

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT ( SECTION -E) Assertion - Reason Type Questions|10 Videos
  • STRUCTURE OF ATOM

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT ( SECTION -F) Matrix - Match Type Questions|4 Videos
  • STRUCTURE OF ATOM

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT ( SECTION -C) Objective Type Questions (More than one options are correct)|15 Videos
  • STRUCTURE OF ATOM

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT (SECTION -D) Assertion-Reason Type Questions|15 Videos
  • SURFACE CHEMISTRY

    AAKASH INSTITUTE ENGLISH|Exercise Assignment Section - C (Assertion - Reason type questions)|10 Videos

Similar Questions

Explore conceptually related problems

The energy of n^(th) orbit is given by E_(n) = ( -Rhc)/(n^(2)) When electron jumpsfrom one orbit to another orbit then wavelength associated with the radiation is given by (1)/(lambda) = RZ^(2)((1)/(n_(1)^(2)) - (1)/ (n_(2)^(2))) When electron of 1.0 gm atom of Hydrogen undergoes transition giving the spectral line of lowest energy in visible region of its atomic spectra, the wavelength of radiation is

The energy of n^(th) orbit is given by E_(n) = ( -Rhc)/(n^(2)) When electron jumps from one orbit to another orbit then wavelength associated with the radiation is given by (1)/(lambda) = RZ^(2)((1)/(n_(1)^(2)) - (1)/ (n_(2)^(2))) The ratio of wavelength H_(alpha) of Lyman Series and H_(alpha) of Pfund Series is

When electron jumps from n = 4 to n = 1 orbit, we get

When an electron jumps from n_(1) th orbit to n_(2) th orbit, the energy radiated is given by

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

Which of the following is associated with the orbital designated by n = 2,l = 1 ?

Which shape is associated with the orbital designated by n = 2,l = 1 ?

If an electron drops from n = 4 to n = 2 in Li^(2+) then would the released wavelength lie in the visible region?

Find the sum to n term of the series whose n^(th) term is given by n(n^(2) + 1)

Energy of the electron in nth orbit of hydrogen atom is given by E_(n) =-(13.6)/(n^(2))eV . The amount of energy needed to transfer electron from first orbit to third orbit is