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Wavelength of first line of lyman series is 1216a°the wavelength of second member of balmer series is ? |
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Answer» ⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐ ___________________________ Hey mate! First member of lyman series(2---1) Also, 1/lembda = RZ^2(1/n1^2 -1/n2^2) Where, R =1/912A° Lembda given =1216A° So, 1/1216A° =1/912A°Z^2(1/1^2 -1/2^2) 3/4 =Z^2(3/4) Z =1 Now, Second member of BALMAR series(4---2) Putting the value of Z =1 In the above formula.. You will get.. Lembda =4864A° Here..n1=2 And n2=4 Hope it will HELP you.. #phoenix |
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