Calculate the wavelength in Angstroms of the photon that is emitted when an electron in the Bohr's orbit `n = 2`, returns to the orbit `n = 1`, in the hydrogen atom .The ionisation potential of the ground state hydrogen atom is `2.17 xx 10^(-11)` ergs per atom

Explanation

प्रश्नानुसार
प्रथम बोर कक्षा में इलेक्ट्रॉन की ऊर्जा `(E_(1))=-2.17xx10^(-11)` अर्ग
अब `E_(n)=(E_(1))/(n^(2))`
अतः `E_(2)=(E_(1))/(2^(2))=(-2.17xx10^(-11))/(2^(2))` अर्ग
`=-(2.17xx10^(-11))/4` अर्ग `=-0.5425xx10^(-11)` अर्ग
`:.DeltAE=E_(2)-E_(1)`
`=(-0.5425xx10^(11))-(-2.17xx10^(-11))`
`=1.6275xx10^(-11)` अर्ग
अब `DeltaE=hv=(hc)/(lamda)`
`:.lamda=(hc)/(DeltaE)=(6.62xx10^(-27)xx3xx10^(10))?(1.6275xx10^(-11))`
`=19.86/1.6275xx10^(-6)\12.25xx10^(-6)` सेमी `=1225Å`