Question

Calculate the two longest wavelength of the radiation emitted when hydrogen atom make transition from higher state to n=2

Answer 1

E=hv

(Where EEE is energy, hhh is Planck’s constant, and vvv is the frequency)

E​7​​−E​6​​=​7​2​​​​−R​H​​​​−​6​2​​​​−R​H​​​​

(simplify by finding a common denominator)

−RH49−−RH36=13RH1764\dfrac{-R_H}{49}\,-\,\dfrac{-R_H}{36}\,=\,\dfrac{13R_H}{1764}​49​​−R​H​​​​−​36​​−R​H​​​​=​1764​​13R​H​​​​

v=​h​​E​​

v=13RH1764hv\,=\,\dfrac{13R_H}{1764h}v=​1764h​​13R​H​​​​

v=131764×2.179×10−18J6.626×10−34J⋅sv\,=\,\dfrac{13}{1764}\times \dfrac{2.179\times 10^{-18}\text{J}}{6.626\times 10^{-34}\text{J}\cdot \text{s}}v=​1764​​13​​×​6.626×10​−34​​J⋅s​​2.179×10​−18​​J​​

v=2.42×1013v\,=\,2.42\times 10^{13}v=2.42×10​13​​ /s

λ=​v​​c​​

(Where λ\lambdaλ is the wavelength, ccc is the speed of light, vvv is the frequency)

λ=2.998×108m/s2.42×1013/s\lambda\,=\,\dfrac{2.998\times 10^8\text{m/s}}{2.42\times 10^{13}/s}λ=​2.42×10​13​​/s​​2.998×10​8​​m/s​​

λ=1.238×10−5\lambda\,=\,1.238\times 10^{-5}λ=1.238×10​−5​​ m

Final Answer:

The longest wavelength is 1.238×10−5 1.238\times 10^{-5}1.238×10​−5​​ m

BeBrainly

Answer 2

Answer:

E=hv

(Where EEE is energy, hhh is Planck’s constant, and vvv is the frequency)

E​7​​−E​6​​=​7​2​​​​−R​H​​​​−​6​2​​​​−R​H​​​​

(simplify by finding a common denominator)

−RH49−−RH36=13RH1764\dfrac{-R_H}{49}\,-\,\dfrac{-R_H}{36}\,=\,\dfrac{13R_H}{1764}​49​​−R​H​​​​−​36​​−R​H​​​​=​1764​​13R​H​​​​

v=​h​​E​​

v=13RH1764hv\,=\,\dfrac{13R_H}{1764h}v=​1764h​​13R​H​​​​

v=131764×2.179×10−18J6.626×10−34J⋅sv\,=\,\dfrac{13}{1764}\times \dfrac{2.179\times 10^{-18}\text{J}}{6.626\times 10^{-34}\text{J}\cdot \text{s}}v=​1764​​13​​×​6.626×10​−34​​J⋅s​​2.179×10​−18​​J​​

v=2.42×1013v\,=\,2.42\times 10^{13}v=2.42×10​13​​ /s

λ=​v​​c​​

(Where λ\lambdaλ is the wavelength, ccc is the speed of light, vvv is the frequency)

λ=2.998×108m/s2.42×1013/s\lambda\,=\,\dfrac{2.998\times 10^8\text{m/s}}{2.42\times 10^{13}/s}λ=​2.42×10​13​​/s​​2.998×10​8​​m/s​​

λ=1.238×10−5\lambda\,=\,1.238\times 10^{-5}λ=1.238×10​−5​​ m

Final Answer:

The longest wavelength is 1.238×10−5 1.238\times 10^{-5}1.238×10​−5​​ m

Explanation: