बामर श्रेणी की उस रेखा के तरंग – दैर्घ्य की गणना करे जो इलेक्ट्रॉन के चौथी कक्षा से दूसरी कक्षा पर कूदने से प्राप्त होती है | [रिडवर्ग स्थिरांक ` ( R ) = 109,677 cm ^ ( – 1 ) `]
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तरंग संख्या ` bar v = 109,677 (( 1 ) / ( n _ 1 ^ 2 ) – ( 1 ) / ( n _ 2 ^ 2 ) ) cm ^ ( -1 )`
प्रश्नानुसार ` n _ 1 = 2 ` और ` n`, अतः
` bar v = 109,677 ( ( 1 ) / ( 2 ^ 2 ) – ( 1 ) /( 4 ^ 2 ) ) cm ^ ( – 1 ) `
` = 20564.4 cm ^ ( – 1 ) `
` because ` तरंग दैर्घ्य ` (bar v) = ( 1 ) /( “तरंग दैर्घ्य ” ( lamda ) ) `
` therefore ` तरंग – दैर्घ्य ` ( lamda ) = ( 1 ) / ( v ) `
` = ( 1 ) / ( 20564.4 ) cm `
` = 0.00004862 cm `
` = 4862 xx 10 ^( -8) cm `
` = 4862 Å ( because 1 Å = 10 ^( – 8 ) cm ) `