Yellow light emitted from a sodium lamp has a wavelength \$\displaystyle \left( \lambda \right) \$ of 580 nm. Calculate the frequency (v) and wavenumber \$\displaystyle \left( \bar { v } \right) \$ of the yellow light.

The wavelength of yellow light is \$\displaystyle 580nm \$ = \$\displaystyle 580 \times 10^{-9}m \$.The frequency of yellow light, \$\displaystyle \nu = \frac {c}{\lambda} = \frac {3.0 \times 10^{8}m/s}{580 \times 10^{-9}m}= 5.17 \times 10^{14}/s\$.The wave number of yellow light, \$\displaystyle \bar \nu = \frac {1}{\lambda} = \frac {1}{580 \times 10^{-9}m}=1.72 \times 10^{6}/m \$.

Solve Study Textbooks Guides

Join / Login

>>Class 11
>>Chemistry
>>Structure of Atom
>> $Development Leading to Bohr's Model of Atom$

Development Leading to Bohr's Model of Atom

Revise with Concepts

Electromagnetic Radiation

ExampleDefinitionsFormulaes

Black Body Radiation

ExampleDefinitionsFormulaes

Photoelectric Effect

ExampleDefinitionsFormulaes

Dual Behaviour of Electromagnetic Radiation

ExampleDefinitionsFormulaes

Learn with Videos

Q: The work function of caesium metal is 2.14 eV. When the light of frequency \$6\times 10^{14}\, Hz\$ is incident on the metal surface, photoemission of electrons occurs. Calculate:(a) the maximum kinetic energy of the emitted electrons,(b) stopping potential,(c) the maximum speed of the emitted photoelectrons?

(a). Work function, \$\phi_o=2.14eV\$Frequency of light, \$\nu=6\times 10^{14}Hz\$Maximum K.E. is given by the photoelectric effect.\$K=h\nu-\phi_o\$\$K=\left(\dfrac{6.626\times 10^{-34}\times 6\times 10^{14}}{1.6\times 10^{-19}}\right)-2.14=0.345eV\$(b). For stopping potential, \$V_o\$, we can write the equation for kinetic energy as,\$K=eV_o\Rightarrow V_o=\dfrac{K}{e}\$\$=\dfrac{0.345\times 1.6\times 10^{-19}}{1.6\times 10^{-19}}=0.345V \$(c). The maximum speed of the emitted photoelectrons is \$v\$.Hence, the relation for kinetic energy can be written as,\$K=\dfrac{1}{2}mv^2\$\$m=9.1\times 10^{-31}kg\$ is the mass of electron.\$\Rightarrow v^2=2K/m\$\$\Rightarrow v=3.323\times 10^5m/s=332.3km/s\$

Q: Calculate energy of one mole of photons of radiation whose frequency is \$5 \times { 10 }^{ 14 }.\$

The energy of one photon (E) \$=hv\$Here, \$h=6.626\times {10^{34}}\$ js \$v= 5\times 10^{14} s^{-1}\$\$E=6.626\times 10^{-34} \times 5\times 10^{14}\$\$3.313\times 10^{-19}\$ jThus the energy of one mole of photons is \$E\times A\$ Where A is the Avogadro number\$=3.313\times 10^{-19}j\times 6.022\times 10^{-23}\$ mol\$=199.50 \times 10^{-3}j mol\$

Q: Calculate (a) frequency and (b) wave-number of yellow radiation having wavelength \$5800 \overset{o} A\$.

Wavelength of light \$ = 5800\times 10^{-10}m \$frequency \$ = \frac{C}{wavelength} = \frac{3\times 10^{8}}{5800\times 10^{-10}} = 0.05\times 10^{16} \$wave no.is reciprocal of wavelength wave no. = \$ \frac{1}{wavelength} = 0.017\times 10^{8}m^{-1} \$

Q: A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.

Power of the bulb = \$100\$ watt = \$100Js^{−1}\$Energy of one photon is \$E=h\nu=hc/\lambda\$where, \$h=6.626 \times 10^{−34}Js, c=3 \times 10^{8}ms^{-1}\$Given \$\lambda=400nm=400 \times 10^{-9}m\$By putting the values, we get\$E=6.626\times 10^{−34}\times 3\times 10^8 /( 400 \times 10^{-9})\$\$E= 4.969 \times 10^{−19}J \$Number of photons emitted in 1 sec \$\times\$ energy of one photon = power\$n \times 4.969 \times 10^{−19}=100\$\$n=\dfrac{100}{4.969 \times 10^{−19}}\$\$n=2.012 \times 10^{20}\$photons per secSo, option \$C\$ is correct

Q: The value of Planck's constant is \$6.63\times10^{34}\$ Js. The speed of light is \$3\times10^{17}\$ nm \$s^{1}\$. Which value is closest to the wavelength in nanometer of a quantum of light with a frequency of \$6\times10^{15}s^{1}\$?

Given:-Planck's constant \$h=6.63\times { 10 }^{ -34 }\ Js\$Speed of light \$c=3\times { 10 }^{ 17 }\ nm{ s }^{ -1 }\$Frequency of quanta \$\nu =6\times { 10 }^{ 15 }{ s }^{ -1 }\$Wavelength, \$\lambda=?\$We know that \$h=\cfrac { c }{ \lambda } \$or \$\lambda =\cfrac { c }{ \nu } \$\$\lambda=\cfrac { 3\times { 10 }^{ 17 }nm{ s }^{ -1 } }{ 6\times { 10 }^{ 15 }{ s }^{ -1 } } \$\$\lambda=50\ nm\$So, option C is correct.

Q: Calculate the mass of a photon with wavelength 3.6\${ A }^{ 0 }\$. Name the principle/equation/concept used. (\$1{ A }^{ 0 }=1\times { 10 }^{ -10 }m\$, speed of light\$=3\times { 10 }^{ 8 }{ ms }^{ -1 }\$, \$h=6.626\times { 10 }^{ -34 }JS\$) 03

\$ \dfrac{hc}{\lambda }=mc^{2}\$ \$m=\dfrac{h}{\lambda c}\$ \$=\dfrac{6.626\times 10^{-34}}{3.6\times 10^{-10}\times 3\times 10^{8}}\$ \$=\dfrac{6.626}{3.6\times 3}\times 10^{-32}\$ \$=0.61\times 10^{-32}kg\$

Q: A gas absorbs a photon of \$355 nm\$ and emits two wavelengths. If one of the emission is at \$680 nm\$, the other is at:

Absorbed energy \$=\$ Emitted energy\$\dfrac{hc}{\lambda} = \dfrac{hc}{{\lambda}_{1}} + \dfrac{hc}{{\lambda}_{2}}\$\$\dfrac{1}{\lambda} = \dfrac{1}{{\lambda}_{1}} + \dfrac{1}{{\lambda}_{2}}\$\$\dfrac{1}{355} = \dfrac{1}{680} + \dfrac{1}{{\lambda}_{2}}\$\${\lambda}_{2}=743 nm\$.

Q: Calculate the wavelength, frequency and wave number of a light wave whose period is \$\displaystyle 2.0\times { 10 }^{ -10 }\ s\$.

(i) The frequency of the light wave is \$\displaystyle \frac {1}{period} = \frac {1}{2.0 \times 10^{-10}s}=5 \times 10^{9}/s\$.(ii) The wavelength of light wave is \$\displaystyle \frac {c}{\nu} = \frac {3 \times 10^{8}m/s}{5 \times 10^{9}/s} = 6.0 \times 10^{-2}m\$.(iii) The wave number of light wave is \$\displaystyle \bar \nu = \frac {1}{\lambda} = \frac {1}{6.0 \times 10^{-2}} =16.66/m\$.

Q: Electromagnetic radiation of wavelength \$242\ nm\$ is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in \$\displaystyle kJ{ mol }^{ -1 }\$.

The ionization energy of sodium will be equal to the energy possessed by the radiation of wavelength 242 nm. It is \$\displaystyle E = \frac {hc}{\lambda} = \frac {6.626 \times 10^{-34}Js \times 3 \times 10^{8} }{242 \times 10^{-9}m } = 8.21 \times 10^{-19}J. \$The ionization energy per mole is \$\displaystyle 8.21 \times 10^{-19} \times 6.023 \times 10^{23} = 494000J =494kJ \$.

Dual Nature of Electromagnetic Radiation

4 mins

Concept behind Electromagnetic Wave Theory

8 mins

Electromagnetic Spectrum - II

4 mins

Electromagnetic Spectrum - I

8 mins

Maxwell's Electromagnetic Theory

7 mins

Characteristics of Electromagnetic Waves

7 mins

Electromagnetic Waves

7 mins

Wave motion and its characteristics

Numericals on Photoelectric effect

12 mins

Quick Summary With Stories

Dual behaviour of matter

2 mins read

View solution

Electromagnetic radiation of wavelength $242 n m$ is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in $k J m o l^{- 1}$ .

Medium

View solution

Revise with Concepts

Electromagnetic Radiation

Black Body Radiation

Photoelectric Effect

Dual Behaviour of Electromagnetic Radiation

Learn with Videos

Quick Summary With Stories

Dual behaviour of matter

Blackbody Radiation

The work function of caesium metal is 2.14 eV. When the light of frequency 6×1014Hz is incident on the metal surface, photoemission of electrons occurs. Calculate: (a) the maximum kinetic energy of the emitted electrons,(b) stopping potential,(c) the maximum speed of the emitted photoelectrons?

Calculate energy of one mole of photons of radiation whose frequency is 5×1014.

Calculate (a) frequency and (b) wave-number of yellow radiation having wavelength 5800Ao.

A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.

The value of Planck's constant is 6.63×1034 Js. The speed of light is 3×1017 nm s1. Which value is closest to the wavelength in nanometer of a quantum of light with a frequency of 6×1015s1?

Calculate the mass of a photon with wavelength 3.6A0. Name the principle/equation/concept used. (1A0=1×10−10m, speed of light=3×108ms−1, h=6.626×10−34JS) 03

A gas absorbs a photon of 355nm and emits two wavelengths. If one of the emission is at 680nm, the other is at:

Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0×10−10 s.

Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (v) and wavenumber (vˉ) of the yellow light.

Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJmol−1.

Calculate energy of one mole of photons of radiation whose frequency is $5 \times 10^{14} .$

Calculate (a) frequency and (b) wave-number of yellow radiation having wavelength $5800 A o$ .

The value of Planck's constant is $6.63 \times 1 0^{34}$ Js. The speed of light is $3 \times 1 0^{17}$ nm $s^{1}$ . Which value is closest to the wavelength in nanometer of a quantum of light with a frequency of $6 \times 1 0^{15} s^{1}$ ?

Calculate the mass of a photon with wavelength 3.6 $A^{0}$ . Name the principle/equation/concept used. ( $1 A^{0} = 1 \times 10^{- 10} m$ , speed of light $= 3 \times 10^{8} m s^{- 1}$ , $h = 6.626 \times 10^{- 34} J S$ ) 03

A gas absorbs a photon of $355 n m$ and emits two wavelengths. If one of the emission is at $680 n m$ , the other is at:

Calculate the wavelength, frequency and wave number of a light wave whose period is $2.0 \times 10^{- 10} s$ .

Yellow light emitted from a sodium lamp has a wavelength $(λ)$ of 580 nm. Calculate the frequency (v) and wavenumber $(\overset{v}{ˉ})$ of the yellow light.

Electromagnetic radiation of wavelength $242 n m$ is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in $k J m o l^{- 1}$ .