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zaharov [31]
3 years ago
9

HEEELLPP!! SEND HELP!! Sometimes goals need to be changed. What are some possible reasons for this?

Physics
1 answer:
kow [346]3 years ago
8 0
Sometimes goals become less important overtime. The older you get, the more your interests and goals would differ from when you were 5. An example is, when I was 5, my goal in life was to stuff candy in my mouth, whereas now, my goal is to have a successful career.
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What is the wavelength, in nm, of the line in the hydrogen spectrum when one n value is 3 and the other n value is 6?
iren [92.7K]

Answer:

\lambda=1090nm

Explanation:

Rydberg formula is used to calculate the wavelengths of the spectral lines of many chemical elements. For the hydrogen, is defined as:

\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})

Where R_H is the Rydberg constant for hydrogen and n_1, n_2 are the lower energy state and the higher energy state, respectively.

\frac{1}{\lambda}=1.10*10^{7}m^{-1}(\frac{1}{3^2}-\frac{1}{6^2})\\\frac{1}{\lambda}=9.17*10^{5}m^{-1}\\\lambda=\frac{1}{1.09*10^{6}m^{-1}}\\\lambda=1.09*10^{-6}m*\frac{10^{9}nm}{1m}\\\lambda=1090nm

4 0
3 years ago
Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2
igomit [66]

Answer

Applying Wein's displacement

\lamda_{max}\ T = 2898 \mu_mK

1) for sun T = 5800 K

      \lambda_{max} = \dfrac{2898}{5800}

      \lambda_{max} = 0.5 \mu_m

2) for tungsten T = 2500 K

      \lambda_{max} = \dfrac{2898}{2500}

      \lambda_{max} = 1.16 \mu_m

3) for heated metal T = 1500 K

      \lambda_{max} = \dfrac{2898}{1500}

      \lambda_{max} = 1.93 \mu_m

4) for human skin T = 305 K

      \lambda_{max} = \dfrac{2898}{305}

      \lambda_{max} = 9.50 \mu_m

5)  for cryogenically cooled metal T = 60 K

      \lambda_{max} = \dfrac{2898}{60}

      \lambda_{max} = 48.3 \mu_m

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ              0.01           0.4               0.7                 100

λT             58           2320            4060             5.8 x 10⁵

F                0             0.125             0.491                1

fractions

for UV = 0.125  

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509  

8 0
4 years ago
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