Patatiniuh

2022-07-15

Determine the maximum frequency and minimum frequency in a) Lyman series. b)Balmer series.

Gornil2

Beginner2022-07-16Added 20 answers

Solution:

According to j.j. blamer formula frequency for hydrogen spectrum is

$d=cR(\frac{1}{{n}_{1}^{2}}+\frac{1}{{n}_{2}^{2}})$

Where,

$R=1.09678\times {10}^{7}{m}^{-1}\phantom{\rule{0ex}{0ex}}c=3\times {10}^{8}m/s$

Maximum frquency:

a) Layman

$f=3\times {10}^{8}\times 1.09678\times {10}^{7}(\frac{1}{{1}^{2}}-\frac{1}{\mathrm{\infty}})\phantom{\rule{0ex}{0ex}}f=3.29\times {10}^{15}\text{}Hz$

Miminum frequency:

$f=3\times {10}^{8}\times 1.09678\times {10}^{7}(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}})\phantom{\rule{0ex}{0ex}}f=2.4675\times {10}^{15}\text{}Hz$

According to j.j. blamer formula frequency for hydrogen spectrum is

$d=cR(\frac{1}{{n}_{1}^{2}}+\frac{1}{{n}_{2}^{2}})$

Where,

$R=1.09678\times {10}^{7}{m}^{-1}\phantom{\rule{0ex}{0ex}}c=3\times {10}^{8}m/s$

Maximum frquency:

a) Layman

$f=3\times {10}^{8}\times 1.09678\times {10}^{7}(\frac{1}{{1}^{2}}-\frac{1}{\mathrm{\infty}})\phantom{\rule{0ex}{0ex}}f=3.29\times {10}^{15}\text{}Hz$

Miminum frequency:

$f=3\times {10}^{8}\times 1.09678\times {10}^{7}(\frac{1}{{1}^{2}}-\frac{1}{{2}^{2}})\phantom{\rule{0ex}{0ex}}f=2.4675\times {10}^{15}\text{}Hz$

bandikizaui

Beginner2022-07-17Added 7 answers

Solution:

According to j.j. blamer formula frequency for hydrogen spectrum is

$d=cR(\frac{1}{{n}_{1}^{2}}+\frac{1}{{n}_{2}^{2}})$

Where,

$R=1.09678\times {10}^{7}{m}^{-1}\phantom{\rule{0ex}{0ex}}c=3\times {10}^{8}m/s$

Maximum frquency:$f=3\times {10}^{8}\times 1.09678\times {10}^{7}(\frac{1}{{2}^{2}}-\frac{1}{\mathrm{\infty}})\phantom{\rule{0ex}{0ex}}f=8.2\times {10}^{15}\text{}Hz$

Minimum frequency:

$f=3\times {10}^{8}\times 1.09678\times {10}^{7}(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}})\phantom{\rule{0ex}{0ex}}f=4.57\times {10}^{15}\text{}Hz$

According to j.j. blamer formula frequency for hydrogen spectrum is

$d=cR(\frac{1}{{n}_{1}^{2}}+\frac{1}{{n}_{2}^{2}})$

Where,

$R=1.09678\times {10}^{7}{m}^{-1}\phantom{\rule{0ex}{0ex}}c=3\times {10}^{8}m/s$

Maximum frquency:$f=3\times {10}^{8}\times 1.09678\times {10}^{7}(\frac{1}{{2}^{2}}-\frac{1}{\mathrm{\infty}})\phantom{\rule{0ex}{0ex}}f=8.2\times {10}^{15}\text{}Hz$

Minimum frequency:

$f=3\times {10}^{8}\times 1.09678\times {10}^{7}(\frac{1}{{2}^{2}}-\frac{1}{{3}^{2}})\phantom{\rule{0ex}{0ex}}f=4.57\times {10}^{15}\text{}Hz$

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