4 Item 4 Learning Goal: To calculate the wavelengths of the lines in the hydrogen emission…
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Question “4 Item 4 Learning Goal: To calculate the wavelengths of the lines in the hydrogen emission…”
Item 4 Learning Goal: To calculate the wavelengths of the lines in the hydrogen emission spectrum Atoms give off light when heated or otherwise excited! The light emitted by excited atoms consists of only a few wavelengths, rather than a full rainbow of colors. When this light is passed through a prism, the result is a series of discrete lines separated by blank areas. The visible lines in the series of the hydrogen spectrum are caused by emission of energy accompanying the fall of an electron from the outer shells to the second shell. The wavelength (X) of the lines can be calculated using the Balmer-Rydberg equation Figure 1 of 1 TTTTTTT 650 700 nm
<Chapter 7 Item 4 a series of discrete lines separated by blank areas. The visible lines in the series of the hydrogen spectrum are caused by emission of energy accompanying the fall of an electron from the outer shells to the second shell. The wavelength (2) of the lines can be calculated using the Balmer-Rydberg equation =R(5- ) where m is an integer, n is an integer greater than m. and R= 1.097 x 10 ? nm is the Rydberg constant Figure 1 of 1 H 40X45 TUTTI 600 650 700 nm
4 08 10 A Review Constants Periodic Table Part A In the Balmer-Rydberg equation (given in the introduction), what value of m is used to determine the wavelengths of the Balmer series? Express your answer as an integer. ► View Available Hint(s) m = Submit
a Review | Constants Periodic Table The image (Figure 1) shows the wavelengths (in nanometers) of the four visible lines in the Balmer series for hydrogen. Match each line to its corresponding transition Drag each item to the appropriate bin. ► View Available Hint(s) Reset Help Indigo line (410 nm) Red line (656 nm) Blue lineBlue-green line (434 nm) (486 nm) n = 6 m – 2 m = 2
(2) 4 of 10 6 A Review | Constants Periodic Table Part 6 If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series emission spectrum. Calculate the wavelength associated with the fifth line Express your answer with the appropriate units. ► View Available Hint(s) = Value Units Submit
Answer
U bene voor I m R = 100 97 X102 nm =2] for Balmer series n=3 &m=2 = 1,097 x162 -1) A = 6560m) B). * n=4 &m=2 I=1.097X102 nm- = 48 6nm l a n=5 &m=2 1 =1.097 xlo 21-116=ualul on=6 &m=2 t = 1.097 xlo2 (1-1) 7=ulonm | 7 nmi
r lms 5th line; (m=2), n=7 T= 1.097xlo? Thus, wandength for 5th line is 3970m o nome tu ya) 314
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