Answer :
Answer:
(a). The first wavelength is 401.0 nm.
(b). The metal's work function is 2.55 eV.
Explanation:
Given that,
Maximum kinetic energy = 0.65 eV
Second wavelength [tex]\lambda_{2}= \dfrac{2}{3}\times\lambda_{1}[/tex]
(a). We need to calculate the wavelength
Using equation of work function for first wavelength
[tex]\dfrac{hc}{\lambda_{1}}-W_{0}=0.65\ eV[/tex].....(I)
For second wavelength,
[tex]\dfrac{hc}{\lambda_{2}}-W_{0}=2.3\ eV[/tex]
Put the value of second wavelength
[tex]\dfrac{1.5 hc}{\lambda_{1}}-W_{0}=2.3\ eV[/tex]....(II)
By subtraction equation (I) from (II)
[tex]0.5\dfrac{hc}{\lambda_{1}}=1.55[/tex]
[tex]\lambda_{1}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{3.1\times1.6\times10^{-19}}[/tex]
[tex]\lambda_{1}=4.010\times10^{-7}\ m[/tex]
[tex]\lambda_{1}=401.0\times10^{-9}\ m[/tex]
[tex]\lambda_{1}=401.0\ nm[/tex]
(b). We need to calculate the work function
Using formula of work function
[tex]W_{0}=(\dfrac{hc}{\lambda}-0.55)[/tex]
Put the value into the formula
[tex]W_{0}=(\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1.6\times10^{-19}\times401.0\times10^{-9}}-0.55)[/tex]
[tex]W_{0}=2.55\ eV[/tex]
Hence, (a). The first wavelength is 401.0 nm.
(b). The metal's work function is 2.55 eV.
Answer:
(a) 3.77 x 10^-7 m
(b) 2.65 eV
Explanation:
According to Einstein photoelectric effect equation
[tex]\frac{hc}{\lambda }-W=KE[/tex]
where, W is the work function and KE is the kinetic energy and λ be the wavelength.
(a) For first wavelength
[tex]\frac{6.63\times 10^{-34}\times 3\times 10^8}{\lambda }-W=0.65\times 1.6\times 10^{-19}[/tex]
[tex]\frac{1.989\times 10^{-25}}{\lambda }-W=1.04 \times 10^{-19}[/tex]... (1)
For second wavelength
λ' = 2λ/3
[tex]\frac{6.63\times 10^{-34}\times 3\times 10^8\times 3}{2\times \lambda }-W=2.3\times 1.6\times 10^{-19}[/tex]
[tex]\frac{2.984\times 10^{-25}}{\lambda }-W=3.68 \times 10^{-19}[/tex]... (2)
Subtract equation (1) from equation (2)
[tex]\frac{0.995\times10^{-25}}{\lambda }=2.64\times 10^{-19}[/tex]
λ = 3.77 x 10^-7 m
(b) Put the value of wavelength in equation (1)
[tex]\frac{1.989\times 10^{-25}}{3.77\times10^{-7} }-W=1.04 \times 10^{-19}[/tex]
W = 4.24 x 10^-19 J
W = 2.65 eV