Hydrogen-like ion has the wavelength difference between the first lines of Balmer and lyman series equal to 33.4 nm. The atomic number of that ion equal to
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● Answer - 4
◆ Explaination-
Wavelength in balmer series is given by,
λb = 36 / (5RZ^2)
Wavelength in lyman series is given by,
λl = 4 / (3RZ^2)
Difference is wavelength is,
λb - λl = 36/(5RZ^2) - 4/(3RZ^2)
33.4×10^-7 = (36/5-4/3)/(RZ^2)
33.4×10^-7 = 5.867 / (RZ^2)
Z^2 = 5.867×10^7 / (33.4×1.1×10^5)
Z^2 = 15.97
Z = 3.99 ~ aprox 4
Atomic number of the ion is 4.
Hope this is helpful...
● Answer - 4
◆ Explaination-
Wavelength in balmer series is given by,
λb = 36 / (5RZ^2)
Wavelength in lyman series is given by,
λl = 4 / (3RZ^2)
Difference is wavelength is,
λb - λl = 36/(5RZ^2) - 4/(3RZ^2)
33.4×10^-7 = (36/5-4/3)/(RZ^2)
33.4×10^-7 = 5.867 / (RZ^2)
Z^2 = 5.867×10^7 / (33.4×1.1×10^5)
Z^2 = 15.97
Z = 3.99 ~ aprox 4
Atomic number of the ion is 4.
Hope this is helpful...
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