Light of wavelength 3000Å falls on a metal surface having work function \(2.3 eV\). Calculate the maximum velocity of ejected electrons. (Planck's constant \(h=6.63 \times 10^{34}\) J.s., Velocity of light \(c=3 \times 10^8 m / s\), Mass of an electron \(=9.1 \times 10^{-31} kg\) )
(July 2016)
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Given:$\lambda = 3000 Å = 3 x 10^{–7} m, m_e = 9.1 10^{–31} kg,$
$Φo = 2.3 eV, h = 6.63 10^{–34} Js, c = 3 10^8 m/s$
To find: Maximum velocity $(v_{max})$
Formula: \(( K . E )_{\max }=\frac{h c}{\lambda}-\Phi_o\)
Calculation: From formula,
\((\text { K.E. })_{\max }=\left[\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3 \times 10^{-7}}-\left(2.3 \times 1.6 \times 10^{-19}\right)\right] J\)
\(=\left(6.63 \times 10^{-19}\right)-\left(3.68 \times 10^{-19}\right) J\)
\(=2.95 \times 10^{-19} J\)
Also,
\(\frac{1}{2} m _{ e } v _{\max }^2=( K . E .)_{\max }\)
\(\therefore \quad v_{\max }=\sqrt{\frac{2(K . E)_{\max }}{m_e}}\)
\(=\sqrt{\frac{2 \times 2.95 \times 10^{-19}}{9.1 \times 10^{-31}}}\)
\(=\sqrt{\frac{5.90 \times 10^{12}}{9.1}}\)
\(=\sqrt{\frac{590}{9 \cdot 1}} \times 10^5\)
\(=\left\{\operatorname{antilog}\left\{\frac{1}{2}[\log (590)-\log (9 \cdot 1)]\right\}\right\} \times 10^5\)
\(=\left\{\operatorname{antilog}\left[\frac{1}{2}(2 \cdot 7709-0.9590)\right]\right\} \times 10^5\)
\(=\left\{\right.\) antilog \(\left.\left[\frac{1}{2}(1.8119)\right]\right\} \times 10^5\)
\(=\{\operatorname{antilog}(0.9059)\} \times 10^5\)
$= 8.052 x 10^5 m/s$
The maximum velocity of electron is $8.052 x 10^5 m/s$
$Φo = 2.3 eV, h = 6.63 10^{–34} Js, c = 3 10^8 m/s$
To find: Maximum velocity $(v_{max})$
Formula: \(( K . E )_{\max }=\frac{h c}{\lambda}-\Phi_o\)
Calculation: From formula,
\((\text { K.E. })_{\max }=\left[\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3 \times 10^{-7}}-\left(2.3 \times 1.6 \times 10^{-19}\right)\right] J\)
\(=\left(6.63 \times 10^{-19}\right)-\left(3.68 \times 10^{-19}\right) J\)
\(=2.95 \times 10^{-19} J\)
Also,
\(\frac{1}{2} m _{ e } v _{\max }^2=( K . E .)_{\max }\)
\(\therefore \quad v_{\max }=\sqrt{\frac{2(K . E)_{\max }}{m_e}}\)
\(=\sqrt{\frac{2 \times 2.95 \times 10^{-19}}{9.1 \times 10^{-31}}}\)
\(=\sqrt{\frac{5.90 \times 10^{12}}{9.1}}\)
\(=\sqrt{\frac{590}{9 \cdot 1}} \times 10^5\)
\(=\left\{\operatorname{antilog}\left\{\frac{1}{2}[\log (590)-\log (9 \cdot 1)]\right\}\right\} \times 10^5\)
\(=\left\{\operatorname{antilog}\left[\frac{1}{2}(2 \cdot 7709-0.9590)\right]\right\} \times 10^5\)
\(=\left\{\right.\) antilog \(\left.\left[\frac{1}{2}(1.8119)\right]\right\} \times 10^5\)
\(=\{\operatorname{antilog}(0.9059)\} \times 10^5\)
$= 8.052 x 10^5 m/s$
The maximum velocity of electron is $8.052 x 10^5 m/s$
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