238744
The de-Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $\left[v_0\right.$ is threshold frequency]
238748
A photon having a wavelength of $845 \AA$, causes the ionisation of $\mathrm{N}$ atom. What is the ionisation energy of $N$ ?
1 $1.4 \mathrm{~kJ}$
2 $1.4 \times 10^4 \mathrm{~kJ}$
3 $1.4 \times 10^2 \mathrm{~kJ}$
4 $1.4 \times 10^3 \mathrm{~kJ}$
Explanation:
: The ionisation energy is the amount of energy required to take out most loosely bonded electron from isolated gaseous atom. $\therefore$ Ionisation energy of nitrogen $=$ energy of photon $=\mathrm{Nh} \frac{\mathrm{c}}{\lambda}$ Where $$ \begin{aligned} & \mathrm{N}=6.02 \times 10^{23} \\ & \mathrm{c}=3 \times 10^8 \\ & \lambda=854 \AA=854 \times 10^{-10} \mathrm{M} \\ & =\frac{6.02 \times 10^{23} \times 6.6 \times 10^{-34} \times 3 \times 10^8}{854 \times 10^{-10}} \\ & =1.4 \times 10^6 \mathrm{~J} \mathrm{~mol}^{-1}=1.4 \times 10^3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \end{aligned} $$
BCECE-2003
Structure of Atom
238749
The increasing order of wavelength for $\mathrm{He}^{+}$ion, neutron (n) and electron (e) particles, moving with the same velocity is-
238744
The de-Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $\left[v_0\right.$ is threshold frequency]
238748
A photon having a wavelength of $845 \AA$, causes the ionisation of $\mathrm{N}$ atom. What is the ionisation energy of $N$ ?
1 $1.4 \mathrm{~kJ}$
2 $1.4 \times 10^4 \mathrm{~kJ}$
3 $1.4 \times 10^2 \mathrm{~kJ}$
4 $1.4 \times 10^3 \mathrm{~kJ}$
Explanation:
: The ionisation energy is the amount of energy required to take out most loosely bonded electron from isolated gaseous atom. $\therefore$ Ionisation energy of nitrogen $=$ energy of photon $=\mathrm{Nh} \frac{\mathrm{c}}{\lambda}$ Where $$ \begin{aligned} & \mathrm{N}=6.02 \times 10^{23} \\ & \mathrm{c}=3 \times 10^8 \\ & \lambda=854 \AA=854 \times 10^{-10} \mathrm{M} \\ & =\frac{6.02 \times 10^{23} \times 6.6 \times 10^{-34} \times 3 \times 10^8}{854 \times 10^{-10}} \\ & =1.4 \times 10^6 \mathrm{~J} \mathrm{~mol}^{-1}=1.4 \times 10^3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \end{aligned} $$
BCECE-2003
Structure of Atom
238749
The increasing order of wavelength for $\mathrm{He}^{+}$ion, neutron (n) and electron (e) particles, moving with the same velocity is-
238744
The de-Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $\left[v_0\right.$ is threshold frequency]
238748
A photon having a wavelength of $845 \AA$, causes the ionisation of $\mathrm{N}$ atom. What is the ionisation energy of $N$ ?
1 $1.4 \mathrm{~kJ}$
2 $1.4 \times 10^4 \mathrm{~kJ}$
3 $1.4 \times 10^2 \mathrm{~kJ}$
4 $1.4 \times 10^3 \mathrm{~kJ}$
Explanation:
: The ionisation energy is the amount of energy required to take out most loosely bonded electron from isolated gaseous atom. $\therefore$ Ionisation energy of nitrogen $=$ energy of photon $=\mathrm{Nh} \frac{\mathrm{c}}{\lambda}$ Where $$ \begin{aligned} & \mathrm{N}=6.02 \times 10^{23} \\ & \mathrm{c}=3 \times 10^8 \\ & \lambda=854 \AA=854 \times 10^{-10} \mathrm{M} \\ & =\frac{6.02 \times 10^{23} \times 6.6 \times 10^{-34} \times 3 \times 10^8}{854 \times 10^{-10}} \\ & =1.4 \times 10^6 \mathrm{~J} \mathrm{~mol}^{-1}=1.4 \times 10^3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \end{aligned} $$
BCECE-2003
Structure of Atom
238749
The increasing order of wavelength for $\mathrm{He}^{+}$ion, neutron (n) and electron (e) particles, moving with the same velocity is-
238744
The de-Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $\left[v_0\right.$ is threshold frequency]
238748
A photon having a wavelength of $845 \AA$, causes the ionisation of $\mathrm{N}$ atom. What is the ionisation energy of $N$ ?
1 $1.4 \mathrm{~kJ}$
2 $1.4 \times 10^4 \mathrm{~kJ}$
3 $1.4 \times 10^2 \mathrm{~kJ}$
4 $1.4 \times 10^3 \mathrm{~kJ}$
Explanation:
: The ionisation energy is the amount of energy required to take out most loosely bonded electron from isolated gaseous atom. $\therefore$ Ionisation energy of nitrogen $=$ energy of photon $=\mathrm{Nh} \frac{\mathrm{c}}{\lambda}$ Where $$ \begin{aligned} & \mathrm{N}=6.02 \times 10^{23} \\ & \mathrm{c}=3 \times 10^8 \\ & \lambda=854 \AA=854 \times 10^{-10} \mathrm{M} \\ & =\frac{6.02 \times 10^{23} \times 6.6 \times 10^{-34} \times 3 \times 10^8}{854 \times 10^{-10}} \\ & =1.4 \times 10^6 \mathrm{~J} \mathrm{~mol}^{-1}=1.4 \times 10^3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \end{aligned} $$
BCECE-2003
Structure of Atom
238749
The increasing order of wavelength for $\mathrm{He}^{+}$ion, neutron (n) and electron (e) particles, moving with the same velocity is-
238744
The de-Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $\left[v_0\right.$ is threshold frequency]
238748
A photon having a wavelength of $845 \AA$, causes the ionisation of $\mathrm{N}$ atom. What is the ionisation energy of $N$ ?
1 $1.4 \mathrm{~kJ}$
2 $1.4 \times 10^4 \mathrm{~kJ}$
3 $1.4 \times 10^2 \mathrm{~kJ}$
4 $1.4 \times 10^3 \mathrm{~kJ}$
Explanation:
: The ionisation energy is the amount of energy required to take out most loosely bonded electron from isolated gaseous atom. $\therefore$ Ionisation energy of nitrogen $=$ energy of photon $=\mathrm{Nh} \frac{\mathrm{c}}{\lambda}$ Where $$ \begin{aligned} & \mathrm{N}=6.02 \times 10^{23} \\ & \mathrm{c}=3 \times 10^8 \\ & \lambda=854 \AA=854 \times 10^{-10} \mathrm{M} \\ & =\frac{6.02 \times 10^{23} \times 6.6 \times 10^{-34} \times 3 \times 10^8}{854 \times 10^{-10}} \\ & =1.4 \times 10^6 \mathrm{~J} \mathrm{~mol}^{-1}=1.4 \times 10^3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \end{aligned} $$
BCECE-2003
Structure of Atom
238749
The increasing order of wavelength for $\mathrm{He}^{+}$ion, neutron (n) and electron (e) particles, moving with the same velocity is-
