What is the wavelength of light emitted when the electron in hydrogen atom undergoes transition from an energy level with n 4 to an energy level with n 2?

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Where λ = Wavelength of the light. – So, the electrons release 486 nm of light when it jumps from higher energy level to lower energy level in the atom.

What is the wavelength of the absorbed photon?

For example, red light has wavelength of 680 nm and contains a considerable amount of energy, 42 kcal. However, a pigment molecule absorbing blue light, with a wavelength 400 nm, absorbs about 71 kcal of energy….Variables:

What are the frequency and wavelength of photon?

Each photon has a wavelength and a frequency. The wavelength is defined as the distance between two peaks of the electric field with the same vector. The frequency of a photon is defined as how many wavelengths a photon propagates each second. Unlike an electromagnetic wave, a photon cannot actually be of a color.

What is the frequency and wavelength of a photon emitted during a transition?

Speed of electromagnetic waves = c = 3 × 108 m/s Planck’s constant = h = 6.63 × 10-34 Js. We want to calculate the frequency, wavelength and energy of the emitted photon, when the atom makes a transition from ni = 4 to nf = 2. = −13.6 × 3 16 eV = 2.55 eV.

What is the energy of a photon of red light that has a wavelength of 656 nm?

That red line has a wavelength of 656 nm. If you do the math you’ll see that the frequency is then 4.57 X 1014 Hz. Using the Planck’s constant equation it can be determined that each photon of this light carries 3.03 X 10-19 J of energy.

What is frequency and wavelength of photon emitted during a transition from N 5 to N 2?

What are frequency and wavelength of a photon emitted during a transition from n=5 state to n = 2 state in hydrogen atom? As n1= 2 and n2 = 5, this transition gives rise to a spectral line in the Balmer series. A photon of wavelength 4 x 10-7 m strikes on metal surface, the work function of the metal being 2.13 eV.

What is photoelectric emission effect?

The photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material. Electrons emitted in this manner are called photoelectrons.

In which region of the electromagnetic spectrum will this radiation lie?

Electromagnetic radiation spectrum is having many types of radiations, which are arranged in a decreasing or increasing order of their wavelengths. Increasing order of wavelength is as: cosmic rays < gamma rays < x-rays < UV radiations < visible light < infrared radiations < microwaves < radio waves.

What is energy of a photon emitted during a transition from n 5 state to the n 2 state in the hydrogen atom?

You can use the fact that a photon emitted during the transition from n = 5 to n = 2 will carry an energy E equal to the difference between the energies of these two states. ( h is Planck’s constant.) where −13.6 eV is the approximate ground-state energy of the hydrogen atom.

What is the shortest wavelength photon it can emit give your answer in nanometers?

The shortest wavelength of a photon that can be emitted by a hydrogen atom, for which the initial state is n = 3, is closest to: The answer is 820 nm, but I don’t know how they got the answer.

How do you find peak amplitude?

To calculate the ‘Peak’ amplitude, we divide the RMS value by 0.707….

1. Technically, the RMS value of a pure sinusoid is equal to the area under the half-wave.
2. That corresponds to (peak x 0.707) on a pure sinusoid.
3. Pure sinusoids are rarely encountered in the world of mechanical vibration.

What is the difference between a peak and amplitude?

Amplitude is the fluctuation or displacement of the wave from its mean value. Peak-to-peak (pk-pk) is the difference between the maximum positive and the maximum negative amplitudes of the wave.

How do you find RMS amplitude?

The RMS amplitude format is calculated by squaring the peak amplitude (A) of the sine wave, diving it by two, and then taking the square root of that quantity. For a single sine wave, the RMS amplitude can be represented as 0.707*A.

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Text Solution

Solution : According to Rydberg equation, <br> `1/lambda = 109678 (1/(n_1^2) - 1/(n_2^2)) cm^(-1)` <br> Here `n_1 = 1, n_2 = 4` <br> `:. 1/lambda = 109678(1/(1^2) - 1/(4^2)) cm^(-1)` <br> `= 109678 xx 15/16 cm^(-1)` <br> or `lambda = (16)/(109678 xx 15 cm^(-1)) = 9.72 xx 10^(-6) cm` <br> `= 97.2 xx 10^(-9) m` <br> `= 97.2 nm`.

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