Question

When sodium is excited in a flame, two ultraviolet spectral lines at lambda - 372.1 nm and lambda = 376.4 nm respectively are emitted . which wavelength represented photons?

a) higher energy?
b) longer wavelengths?
c) higher frequences?

Answer 1

Answer: Wavelength which represented photons with :

a) Higher energy is 372.1 nm

b) Longer wavelength is 376.4 nm

c) Higher frequency is 372.1 nm

Step-by-step explanation:

`E=h\\u =(hc)/(\lamda )` (Planck's equation)

`h=\text{Planck's constant}=6.62* 10^(-34)J-s,c=\text{speed of light}=3* 10^(8)m/s`

`\lambda` = wavelength of the photon with energy E in meters.

`\\u` = frequency of the photon with energy E in hertz.

For first spectral line:

`\lambda _1=372.1 nm=372.1* 10^(-9) m(1nm=1* 10^(-9) m)`

`E_1=(hc)/(\lambda_1)=((6.62* 10^(-34)J-s)(3* 10^(8)m/s))/(372.1* 10^(-9) m)=5.33* 10^(-19)` joules

`\\u _1=(c)/(\lambda _1)=(3* 10^(8)m/s)/(372.1* 10^(-9) m)=0.08` Hertz

For second spectral line:

`\lambda _2=376.4 nm=376.4* 10^(-9) m(1nm=1* 10^(-9) m)`

`E_2=(hc)/(\lambda_2)=((6.62* 10^(-34)J-s)(3* 10^(8)m/s))/(376.4* 10^(-9) m)=5.27* 10^(-19)` joules

`\\u _2=(c)/(\lambda _2)=(3* 10^(8)m/s)/(376.4* 10^(-9) m)=0.07` Hertz

Answer 2

The wavelength that represented photons is at lambda = 376.4 nm.
The wavelength with the higher energy is at lambda - 372.1 nm.
The longer wavelength is, of course, at lambda 376.4 nm.
The wavelength with the higher frequency is 376.4 nm.