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NCERT Class XI Chemistry Structure of Atom Solutions

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Question : 51 of 67
Marks: +1, -0
The work function for caesium atom is 1.9 eV. Calculate (a) the threshold frequency and (b) the threshold wavelength of the radiation. (c) If the caesium elements is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Solution:  
(a) Work function = hυ0h\upsilon_0 = 1.9 eV = 1.9 × 1.602 × 10−19 J10^{-19}\,\mathrm{J} = 3.04 × 10−19 J10^{-19}\,\mathrm{J}
Threshold frequency, (u0)(u_0) = 3.04×10−196.626×10−34\frac{3.04\times10^{-19}}{6.626\times10^{-34}} = 4.59 × 1014 s−110^{14}\,\mathrm{s}^{-1}
(b) Threshold wavelength, (λ0)(\lambda_0) = cv0\frac{c}{v_0} = 3×1084.59×1014\frac{3\times10^{8}}{4.59\times10^{14}} = 5.64 × 10−7 m10^{-7}\,\mathrm{m}
or 654 × 10−9 m10^{-9}\,\mathrm{m} or 654 nm
(c) Now, energy of light, (E) = hcλ\frac{hc}{\lambda} = 6.626×10−34×3×108500×10−9\frac{6.626\times10^{-34}\times3\times10^{8}}{500\times10^{-9}} = 3.98 × 10−19 J10^{-19}\,\mathrm{J}
Kinetic energy of ejected electron = 3.98 × 10−1910^{-19} – 3.04 × 10−1910^{-19} = 9.4 × 10−20 J10^{-20}\,\mathrm{J}
Since K.E. = 12mv2\frac{1}{2}mv^2
⇒ v = 2 K.E.m\sqrt{\frac{2\,\text{K.E.}}{m}} = 2×9.4×10−209.1×10−31\sqrt{\frac{2\times9.4\times10^{-20}}{9.1\times10^{-31}}} = 4.54 × 105 m s−110^{5}\,\mathrm{m}\,\mathrm{s}^{-1}
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