A chemistry problem by Jay Rashamiya

Chemistry Level pending

when electromagnetic radiation of wavelength 300 nm falls on the surface of sodium,electrons are emmited with kinetic energy 1.68 x 10^5 J/mol.what is the minimmum energy needed to remove an electron from sodium in joule.

3.84 x 10^-19 J 9.87 x 10^-19 J 5.16x 10^-16 J 10^-15 J

Jay Rashamiya by Jay Rashamiya , 21, India

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1 solution

Rohit Ner
Jun 2, 2015

Kinetic energy of each electron= 1.68 × 10 5 6.022 × 10 23 = 2.78 × 10 − 19 J \frac { 1.68\times { 10 }^{ 5 } }{ 6.022\times { 10 }^{ 23 } } =2.78\times { 10 }^{ -19 }J
Energy incident on the surface is given by: E = h c λ E=\frac { hc }{ \lambda } where λ \lambda represents the wavelength of incident light, h h represents the Planck's Constant and c c represents the speed of light.
⇒ E = 6.62 × 10 − 34 × 3 × 10 8 300 × 10 − 9 = 6.62 × 10 − 19 \Rightarrow E=\frac { 6.62\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 300\times { 10 }^{ -9 } } =6.62\times { 10 }^{ -19 }
By conservation of energy,
E = W 0 + K . E . E={ W }_{ 0 }+K.E.
where W 0 { W }_{ 0 } is the minimum energy needed to remove the electron from the surface. ⇒ W 0 = E − K . E . = 6.62 × 10 − 19 − 2.78 × 10 − 19 = 3.84 × 10 − 19 J \Rightarrow { W }_{ 0 }=E-K.E.\\=6.62\times { 10 }^{ -19 }-2.78\times { 10 }^{ -19 }=\large\color{#3D99F6}{\boxed{3.84\times { 10 }^{ -19 }J}}

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