Question

Calculate the value of wavelength of hydrogen electron when it jumps from 5th energy level to 2nd energy level

Answer 1

✌❤HEYA_MATE❤✌

The energy transition will be equal to 1.55⋅

10−19J .

So, you know your energy levels to be n = 5 and n = 3. Rydberg's equation will allow you calculate the wavelength of the photon emitted by the electron during this transition

1λ=R⋅

(1n2final−1n2 initial)

, where

λ - the wavelength of the emitted photon;R - Rydberg's constant - 1.0974⋅107m−1 ;n final - the final energy level - in your case equal to 3;n initial - the initial energy level - in your case equal to 5.

So, you've got all you need to solve for λ , so1λ=1.0974⋅

107m−1⋅(132−152)1λ=0.07804.

107m−1⇒λ=1.28⋅

10−6m

Since

E=hcλ , to calculate for the energy of this transition you'll have to multiply Rydberg's equation by h⋅c, where - Planck's constant - 6.626⋅

10−3J ⋅sc - the speed of light -

299,792,458 m/s

So, the transition energy for your particular transition (which is part of the Paschen Series) is E=6.62⋅10−34J⋅s⋅299,792,458m/s1.28⋅10−6m

E=1.55⋅

10−19J

✌❤PIYUSH_SHARMA❤✌

Answer 2

The second energy level is -3.4 eV. Thus it would take E2 − E1 = -3.4 eV − -13.6 eV = 10.2 eV to excite the electron from the ground state to the first excited state. If a photon has more energy than the binding energy of the electron then the photon will free the electron from the atom – ionizing it.